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Integral Transformations Solved Problems

Example.1 Show that the area bounded by a simple closed curve C is given by \(\frac{1}{2} \oint_C\)xdy-ydx and hence find the area of the ellipse x=a cos θ,y= bsinθ, 0 ≤ θ ≤ 2π

Solution:

By Green’s theorem \(\int_C \mathrm{P} d x+\mathrm{Q} d y=\iint_S\left(\frac{\partial \mathrm{Q}}{\partial x}-\frac{\partial \mathrm{P}}{\partial y}\right) d x d y\)

Put \(\mathrm{P}=-y\) and \(\mathrm{Q}=+x \frac{\partial \mathrm{P}}{\partial y}=-1, \frac{\partial \mathrm{Q}}{\partial x}=+1\)

∴ \(\oint_C x d y-y d x=2 \int_S d x d y=2 \mathrm{~A}\) where A is the area of the surface S

∴ \(\frac{1}{2} \int_C x d y-y d x=\mathrm{A}\)

Now for the ellipse x = \(a \cos \theta, y=b \sin \theta\).

Area = \(\frac{1}{2} \oint_C x d y-y d x\)

= \(\frac{1}{2} \int_0^{2 \pi}(a \cos \theta)(b \cos \theta)-(b \sin \theta)(-a \sin \theta) d \theta=\frac{1}{2} a b \int_0^{2 \pi} d \theta=\pi a b\)

Example.2 Evaluate \(\oint_C\)(cos x sin y – xy) dx+ sin x cos y dy, by Green’s theorem where C is the circle x²+y² =1

Solution:

Given

\(\oint_C\)(cos x sin y – xy) dx+ sin x cos y dy

By Green’s theorem, we have \(\int_C \mathrm{P} d x+\mathrm{Q} d y=\iint_S\left(\frac{\partial \mathrm{Q}}{\partial x}-\frac{\partial \mathrm{P}}{\partial y}\right) d x d y\)

Here \(\mathrm{P}=\cos x \sin y-x y, \mathrm{Q}=\sin x \cos y\)

∴ \(\frac{\partial \mathrm{P}}{\partial y}=\cos x \cos y-x, \frac{\partial \mathrm{Q}}{\partial x}=\cos x \cos y\)

for the circle, \(x^2+y^2=1\). Changing to polar coordinates.

x = \(r \cos \theta, y=r \sin \theta, d x d y=r d r d \theta\)

∴ \(\oint_C(\cos x \sin y-x y) d x+\sin x \cos y d y \int_S[\cos x \cos y-(\cos x \cos y-x)] d x d y\)

= \(\iint_S x d x d y=\int_{\theta=0}^{2 \pi} \int_{r=0}^1 r \cos \theta \cdot r d r d \theta=\int_0^{2 \pi}\left[\frac{r^3}{3}\right]_0^1 \cos \theta d \theta=\frac{1}{3} \int_0^{2 \pi} \cos \theta d \theta=\frac{1}{3}[\sin \theta]_0^{2 \pi}=0\)

Practical Examples Of Green’S Theorem In Vector Calculus

Example.3 Verify Green’s theorem in the plane for \(\oint_C\)(3x²-8y²)dx+(4x-6xy) dy where C is the region bounded by y=\(\sqrt{x}\)  and y=x²

Solution:

Here \(\mathrm{P}=3 x^2-8 y^2 \quad \frac{\partial \mathrm{P}}{\partial y}=-16 y \quad \mathrm{Q}=4 y-6 x y \quad \frac{\partial \mathrm{Q}}{\partial x}=-6 y\)

Hence to Green’s theorem

1. \(\int_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= \(\iint_S(-6 y+16 y) d x d y=10 \int_{x=0}^1 \int_{y=x^2}^C y d y d x\)

=\(10 \int_0^1\left[\frac{y^2}{2}\right]_{x^2}^{\sqrt{x}} d x=5 \int_0^1\left(x-x^4\right) d x=\frac{3}{2}\)

2. Verification.

The line integral along C

= Line integral along y = \(x^2\) (from O to A) + line integral along \(y^2=x\) (from A to O) = \(\mathrm{I}_1+\mathrm{I}_2\)

∴ \(\mathrm{I}=\int_{x=0}^1\left[3 x^2-8\left(x^2\right)^2\right] d x+\left[4 x^2-6 x\left(x^2\right)\right] 2 x d x=\int_0^1\left(3 x^2+8 x^2-20 x^4\right) d x=-1\)

⇒ \(\mathrm{I}_2=\int_1^0\left(3 x^2-8 x\right) d x+\left(4 \sqrt{x}-6 x^{3 / 2}\right) \frac{1}{2 \sqrt{x}} d x\)

y = \(\sqrt{x}\)

= \(\int_1^0\left(3 x^2-11 x+2\right) d x=\frac{5}{2}\)

∴ \(\mathrm{I}_1+\mathrm{I}_2=-1+\frac{5}{2}=\frac{3}{2}\)

Hence the theorem is verified.

Example.4 Evaluate by Green’s theorem \(\oint_C\)(y-sinx)+cos x dy where C is the triangle enclosed by the lines x=0,x=\(\frac{\pi}{2}\), πy=2x.

Solution:

Here P = \(y-\sin x, \mathrm{Q}=\cos x \quad \frac{\partial \mathrm{P}}{\partial y}=1, \frac{\partial \mathrm{Q}}{\partial x}=-\sin x\)

Hence by Green’s theorem \(\int_C(y-\sin x) d x+\cos x d y=\iint_S(-1-\sin x) d x d y\)

= \(-\int_{x=0}^{\pi / 2} \int_{y=0}^{2 x / \pi}(1+\sin x) d x d y=-\int_0^{\pi / 2}(1+\sin x) \frac{2 x}{\pi} d x\)

= \(-\frac{2}{\pi} \int_0^{\pi / 2}(x+x \sin x) d x=-\left(\frac{\pi}{4}+\frac{2}{\pi}\right)\)

Example.5 Evaluate by greens theorem \(\oint_C\)(x-cosh y) dx +(y +sin x) dy where C is the rectangle with vertices (0,0),(π,0) (π,1) and (0,1)

Solution:

Here \(\mathrm{P}=x^2-\cosh y, \quad \mathrm{Q}=y+\sin x \quad \frac{\partial \mathrm{P}}{\partial y}=-\sinh y, \quad \frac{\partial \mathrm{Q}}{\partial x}=\cos x\)

By Green’s theorem \(\int_C\left(x^2-\cosh y\right) d x+(y+\sin x) d y=\iint_S(\cos x+\sinh y) d x d y\)

= \(\int_{x=0}^\pi \int_{y=0}^1(\cos x+\sinh y) d x d y=\int_0^\pi[(y \cos x+\cosh y)]_0^1 d x=\int_0^\pi[\cos x+\cosh 1-1] d x\)

= \([\sin x+x \cosh 1-x]_0^\pi=\pi[\cosh 1-1]\)

Examples Of Green’S Theorem In Physics And Engineering

Example.6 Verify Greens theorem in the plane for \(\oint_C\)(xy+y²) dx+x²dy where C is the closed curve of the region bounded by y=x and y=x²

Solution:  Solving the curves y=x and y=x² we get the points  of intersection are (0,0) and (1,1)

On the curve y=x²

The limits for x are 0 to 1

y²= x  ⇒ dy =2x dx

⇒ \(\int_{C_1}\left(x y+y^2\right) d x+x^2 d y\)

= \(\int_{x=0}^1\left(x\left(x^2\right)+x^4\right) d x+x^2 2 x d x\)

= \(\int_0^1\left(x^3+x^4+2 x^3\right) d x=\int_0^1\left(3 x^3+x^4\right) d x\)

= \(3 \frac{x^4}{4}+\frac{x^5}{5}_0^1=\frac{3}{4}+\frac{1}{5}-0=\frac{19}{20}\)

On the curve y=x

The limits of x are \(1 \rightarrow 0\)

y = \(x \Rightarrow d y=d x\)

⇒ \(\left.\oint_C\left(x y+y^2\right) d x+x^2 d y=\int_{x=1}^0\left(x(x)+x^2\right) d x+x^2 d x=\int_1^0 3 x^2=\frac{3 x^3}{3}\right]_1^0=-1\)

Adding (1) and (2) : \(\oint_C\left(x y+y^2\right) d x+x^2 d y=\frac{19}{20}-1=-\frac{1}{20}\)

By Green’s theorem, \(\oint_C P d x+Q d y=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)

R. H. S. \(=\int_{x=0}^1 \int_{y=x^2}^x\left[\frac{\partial}{\partial x}\left(x^2\right)-\frac{\partial}{\partial y}\left(x y+y^2\right)\right] d x d y\)

= \(\int_{x=0}^1 \cdot \int_{y=x^2}^x(2 x-x-2 y) d x d y=\int_{x=0}^1 \int_{y=x^2}^x(x-2 y) d x d y\)

= \(\left.\int_{x=0}^1 x y-y^2\right]_{x^2}^x d x=\int_0^1\left[\left(x^2-x^2\right)-\left(x^3-x^4\right)\right] d x\)

= \(\int_{x=0}^1\left(x^4-x^3\right) d x=\frac{x^5}{5}-\frac{x^4}{4}_0^1=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}=\text { L. H. S. }\)

∴ Green’s theorem is verified.

Integral Transformations Solved Problems

Example.1 Show that \(\int_S\)(axi+byj+czk).N.dS=4\(\frac{\pi}{3}\) (a+b+c) where S is the surface of the sphere x²+y²+z²=1.

Solution:

Here \(\mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k} \quad \text{div} \mathbf{F}=\nabla . \mathbf{F}=\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=a+b+c\)

By Gauss’s Theorem \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\int_V \nabla \cdot \mathbf{F} d \mathbf{V}=\int_V(a+b+c) d \mathbf{V}=(a+b+c) \mathbf{V}=\frac{4 \pi}{3}(a+b+c)\)

V = \(\frac{4 \pi}{3}\), for the given sphere

Example.2 Verify Gauss’s divergence theorem to evaluate (x³-yz)i-2x²yj+zk).NdS over the surface of a cube bounded by the coordinate planes x=y=z=a.

Solution:

Gauss’s theorem states that \(\iint_{\mathrm{S}} \mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z\)

From the problem \(\mathbf{F}_1=x^3-y z, \mathbf{F}_2=-2 x^2 y, \mathbf{F}_3=z\)

⇒ \(\frac{\partial \mathbf{F}_1}{\partial x}=3 x^2, \frac{\partial \mathbf{F}_2}{\partial y}=-2 x^2, \frac{\partial \mathbf{F}_3}{\partial z}=1\)

∴ RHS = \(\iiint_V\left(3 x^2-2 x^2+1\right) d x d y d z\)

= \(\int_0^a \int_0^a \int_0^a\left(x^2+1\right) d x d y d z\)

= \(\int_0^a \int_0^a\left[\frac{x^3}{3}+x\right]_0^a d y d z=\frac{1}{3} a^5+a^3\)

Integral Transformations For Spherical Surfaces Using Divergence Theorem

Verification: Let us calculate the value of \(\int \mathbf{F} \cdot \mathbf{N} d \mathbf{S}\) over the six faces of the cube directly

1. For the face PQAR \(\mathbf{N}=\mathbf{i}, d \mathbf{S}=d z d y\) and x=a

∴ \(\int_{S_1} \mathbf{F} . \mathbf{N} d \mathbf{S}=\iint_{S_1}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] . \mathbf{i} d z d y\)

= \(\int_{z=0}^a \int_{y=0}^a\left(x^3-y z\right) d z d y=\int_{z=0}^a \int_{y=0}^a\left(a^3-y z\right) d z d y=a^3\left[[y]_0^a[z]_0^a\right]-\left[\frac{y^2}{2}\right]_0^a\left[\frac{z^2}{2}\right]_0^a=a^5-\frac{1}{4} a^4\)

2. For the face OBSC \(\mathbf{N}=-\mathbf{i}, d \mathbf{S}=d y d z\) and x=0

∴ \(\int_{S_2} \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{y=0}^a \int_{z=0}^a\left(x^3-y z\right) d y d z=\int_0^a \int_0^a y z d y d z=\left[\frac{y^2}{2}\right]_0^a\left[\frac{z^2}{2}\right]_0^a=\frac{1}{4} a^4\)

because x=0 for this face

3. For the face \(\text{PQBS} \mathbf{N}=\mathbf{j}, d \mathbf{S}=d z d x\) and y=a

∴ \(\int_{S_3} \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{x=0}^a \int_{z=0}^a\left(2 x^2 y\right) d x d z=-2 a \int_0^a \int_0^a x^2 d x d z\)

because y = a for this face

= \(-2 a\left[\frac{x^3}{3}\right]_0^a[z]_0^a=-\frac{2}{3} a^5\)

4. For the face OARC \(\mathbf{N}=-\mathbf{j}, d \mathbf{S}=d z d x\) and y=0

⇒ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{S_4} 2 x^2 y d x d z=0\) because on this face y=0

5. For the face PRCS \(\mathbf{N}=\mathbf{k}, d \mathbf{S}=d x d y\) and z=a

∴ \(\int_S \mathbf{F} . \mathbf{N} d \mathbf{S}=\int_{x=0}^a \int_{y=0}^a z d x d y=a \int_0^a \int_0^a d x d y\)

because z=a on this face

= \(a[x]_0^a[y]_0^a=a^3\)

6. For the face OBQA \(\mathbf{N}=-\mathbf{k}, d \mathbf{S}=d x d y\) and z=0

∴ \(\int_S \mathbf{F} . \mathbf{N} d \mathbf{S}=-\int_{x=0}^a \int_{y=0}^a z d x d y=0\) because on this face z=0

Hence for the total faces \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=a^5-\frac{1}{4} a^4+\frac{1}{4} a^4-\frac{2}{3} a^5+0+a^3=\frac{1}{3} a^5+a^3\)

Example.3 Apply Gauss’s theorem to prove that \(\int_S\)r.N.dS=3V

Solution:  By Gauss’s theorem \(\int_S\)r.N.dS=\(\int_V\) div r dv

div \(\mathbf{r}=\nabla \cdot \mathbf{r}=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \cdot(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=1+1+1=3\)

∴ \(\int_V \text{div} \mathbf{r} d \mathbf{V}=\int_V 3 d \mathbf{V}=3 V\)

Surface Integral Problems With Gauss’S Divergence On Spheres

Example.4 By transforming into a triple integral, evaluate (x³ dy dz+x²y dz dy+x²z dx dy) where S is the closed surface consisting of the cylinder x²+y²=a² and the circular discs z=0 and z=b.

Solution:

Here \(\mathbf{F}_1=x^3, \mathbf{F}_2=x^2 y, \mathbf{F}_3=x^2 z\)

⇒ \(\frac{\partial \mathbf{F}_1}{\partial x}=3 x^2, \frac{\partial \mathbf{F}_2}{\partial y}=x^2, \frac{\partial \mathbf{F}_3}{\partial z}=x^2\)

∴ \(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}=3 x^2+x^2+x^2=5 x^2\)

By Gauss’s theorem \(\iint_{\mathrm{S}} \mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z\)

∴ \(\iint_{\mathrm{S}} x^3 d y d z+x^2 y d z d x+x^2 z d x d y\)

= \(\iiint_V 5 x^2 d x d y d z=5(4) \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} \int_{z=0}^b x^2 d x d y d z\)

= \(20 \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} x^2(b) d x d y=20 b \int_0^a x^2 \sqrt{a^2-x^2} \cdot d x\)

Put x = \(a \sin \theta, d x=a \cos \theta d \theta\)

x = \(0 \Rightarrow \theta=0 \quad x=a \Rightarrow \theta=\pi / 2\)

= \(20 b a^4 \int_0^{\pi / 2} \sin ^2 \theta \cdot \cos ^2 \theta d \theta=20 a^4 b \int_0^{\pi / 2}\left(\sin ^2 \theta-\sin ^4 \theta\right) d \theta\)

= \(20 a^4 b\left[\frac{1}{2} \frac{\pi}{2}-\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=\frac{5}{4} \pi a^4 b\)

Example.5 Prove that ∫f.curl F dV =∫F×f.dS+∫F.curlf dV

Solution:

Consider the vector function \(\mathbf{F} \times \mathbf{f}\)

By Gauss’s theorem \(\int_S(\mathrm{~F} \times \mathrm{f}) \cdot \overline{\mathrm{N}} d \mathrm{~S}=\int_V \text{div}(\mathrm{F} \times \mathrm{f}) d \mathrm{~V}\)

⇒ \(\int_S(\mathrm{~F} \times \mathbf{f}) \cdot \mathrm{N} d \mathrm{~S}=\int_V \text{div}(\mathrm{F} \times \mathbf{f}) d \mathrm{~V}=\int(\mathbf{F} \times \mathbf{f}) \cdot d \mathbf{S}=\int_V \nabla \cdot(\mathbf{F} \times \mathbf{f}) d \mathbf{V}\)

because \(\mathbf{N} d \mathbf{S}=d \mathbf{S}\)

∴ \(\int(F \times \bar{f}) d \overline{\mathrm{S}}=\int_V(\mathrm{f} . \mathrm{curlF}-\mathrm{F} \text { curlf }) d \mathrm{~V}\)

∴ \(\int \mathbf{f} . \text{curl} \mathbf{F} d \mathbf{V}=\int(\mathbf{F} \times \mathbf{f}) \cdot d \mathbf{S}+\int \mathbf{F} . \text{curl} \mathbf{f} d \mathbf{V}\)

Integration Techniques For Surface Of The Sphere Using Gauss’S Theorem

Example.6 Compute \(\int_S\) (ax²+by²+cz²) ds over the sphere x² +y²+z²=1

Solution:

By divergence theorem \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\int_V \nabla \cdot \mathbf{F} d \mathbf{V}\)

Given \(\mathbf{F}. \mathbf{N}=a x^2+b y^2+c z^2\)

Let \(\phi=x^2+y^2+z^2-1\)

∴ Normal vector N to the surface \(\phi\) is \(\nabla \phi=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(x^2+y^2+z^2-1\right)=2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})\)

∴ Unit normal vector \(\mathbf{N}=\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{2 \sqrt{\left(x^2+y^2+z^2\right)}}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

∴ \(\mathbf{F} . \mathbf{N}=\mathbf{F} .(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=a x^2+b y^2+c z^2\)

i.e. \(\mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}\)

∴ \(\nabla . \mathbf{F}=a+b+c\)

Hence by Gauss theorem \(\oint_S\left(a x^2+b y^2+c z^2\right) d \mathbf{S}=\int_V(a+b+c) d \mathbf{V}\)

=(a+b+c) \(\mathbf{V}=\frac{4 \pi}{3}(a+b+c)\) as \(\mathbf{V}=\frac{4}{3} \pi\) being the volume of the sphere of unit radius.

Example.7 By converting the surface integral to volume integral show that ∫\(\int_S\) x³dy dz + y³dz dx+ z³dx dy= \(\frac{12 \pi}{5}\)where S is the surface of the sphere x²+y²+z²=1

Solution:

Consider the sphere \(x^2+y^2+z^2=a^2\) and

⇒ \(\bar{F}=x^3 \bar{i}+y^3 \bar{j}+z^3 \bar{k}=F_1 \bar{i}+F_2 \bar{j}+F_3 \bar{k}\)

div \(\bar{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=3 x^2+3 y^2+3 z^2=3\left(x^2+y^2+z^2\right)=3 r^2\)

where r is the radius of the sphere.

We have by divergence theorem \(\iint_S \bar{F} \cdot \bar{n} d s=\iiint_{d i v} \bar{F} d \nu\)

i.e. \(\iint_{\text {S }} F_1 d y d z+F_2 d z d x+F_3 d x d y=\iiint_V d i v F d v=3 \iiint_S r^2 d v\)

Changing into a spherical coordinator

x = \(r \sin \theta \cos \phi \quad y=r \sin \theta \sin \phi \quad z=r \cos \theta\)

then \(d V=r^2 \sin \theta d r d \theta d \phi\)

L.H.S. = \(3 \int_{\phi=0}^{2 \pi} \int_{\theta=0}^\pi \int_{r=0}^a r^4 \sin \theta d r d \theta d \phi=3 \int_0^{2 \pi} d \phi \int_0^\pi \sin \theta d \theta \int_0^a r^4 d r\)

= \(3[\phi]_0^{2 \pi}[-\cos \theta]_0^\pi\left[\frac{r^5}{5}\right]_0^a=\frac{12}{5} \pi a^5\)

Taking a=1 we get the value as \(\frac{12}{5} \pi\)

Example.8 Find \(\iint_S \overline{\mathbf{F}} \cdot \overline{\mathbf{N}} d \mathrm{~s}\) where \(\overline{\mathrm{F}}\) =2xy\(\overline{\mathrm{i}}\) +yz2\(\overline{\mathrm{j}}\) +xz\(\overline{\mathrm{k}}\) and S is the surface of the parallelopiped formed by x=0,y=0,z=0,x=0,x=2,y=1,z=3.

Solution:

By Gauss’s theorem \(\iint_S \overline{\mathbf{F}} \cdot \overline{\mathbf{N}} d \mathbf{s}=\iiint_V \nabla \cdot \overline{\mathbf{F}} d \mathbf{v}\)

⇒ \(\nabla. \overline{\mathbf{F}}=\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}=2 y+z^2+x\)

⇒ \(\iiint_V(\nabla . \overline{\mathbf{F}}) d \mathbf{v}=\int_{x=0}^2 \int_{y=0}^1 \int_{z=0}^3\left(x+2 y+z^2\right) d x d y d z=\int_{x=0}^2 \int_{y=0}^1\left[x z+2 y z+\frac{z^3}{3}\right]_0^3 d x d y\)

= \(\int_{x=0}^2 \int_{y=0}^1(3 x+6 y+9) d x d y=\int_{x=0}^2\left[3 x y+3 y^2+9 y\right]_{y=0}^1 d x\)

= \(\int_{x=0}^2(3 x+12) d x=\left[\frac{3 x^2}{2}+12 x\right]_0^2=6+24=30\)

Example.9 Valuate by Gauss divergence theorem∫ \(\int_S\)4xz dy dz-y² dz dx+yz x dy where S is the surface of the cube bounded by the planes x=0,x=1, y=0,y=1,z=0,z=1.

Solution:

By divergence theorem \(\int_S \overline{\mathbf{F}} \cdot \mathbf{N} d \mathbf{S}=\int_V d i v \overline{\mathbf{F}} d \mathbf{V}\)

⇒ \(\iint_S \mathbf{f}_1 d y d z+\mathbf{f}_2 d z d x+\mathbf{f}_3 d x d y=\iiint_V\left(\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}\right) d x d y d z\)

From the given problem \(\mathbf{f}_1=4 x z, \mathbf{f}_2=-y^2\) and \(\mathbf{f}_3=y z\)

We get \(\iiint_V\left[\frac{\partial}{\partial x}(4 x z)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}(y z)\right] d x d y d z\)

= \(\iiint_{x=0}^1(4 z-2 y+y) d x d y d z=\int_{x=0}^1 \int_{y=0}^1 \int_{z=0}^1(4 z-y) d z d y d x\)

= \(\int_{x=0}^1 \int_{x=0}^1\left[2 z^2-y z z_{z=0} d y d x=\int_{x=0}^1 \int_{y=0}^1(2-y) d y d x\)

= \(\int_{x=0}^1\left[2 y-\frac{y^2}{2}\right]_0^1 d x=\int_{x=0}^1\left(2-\frac{1}{2}\right) d x\right.\)

= \(\frac{3}{2} \int_0^1 d x=\frac{3}{2}[x]_0^1=\frac{3}{2}\)

Understanding Integral Transformations On Spherical Surfaces

Example.10 If\(\overline{\mathrm{F}}\) =(2x²-3z)\(\overline{\mathrm{i}}\)-2xy\(\overline{\mathrm{j}}\)-4x\(\overline{\mathrm{k}}\) then evaluate \(\int_S\)div\(\overline{\mathrm{F}}\) dv where V is the closed region bounded by the planes x=0, y=0,z=0 and 2x+2y+z=4.

Solution:

div \(\overline{\mathbf{F}}=\nabla \cdot \overline{\mathbf{F}}=\frac{\partial \mathbf{f}_1}{\partial x}+\frac{\partial \mathbf{f}_2}{\partial y}+\frac{\partial \mathbf{f}_3}{\partial z}=4 x-2 x=2 x\)

Limits of z are 0 to 4-(2 x+2 y)

Limits of y are 0 to 2-x

Limits of x are 0 to 2

∴ \(\int_V \nabla \cdot \overline{\mathbf{F}} d \mathbf{V}=\int_{x=0}^2 \int_{y=0}^{2-x} \int_{z=0}^{4-2 x-2 y} 2 x d x d y d z=\int_{x=0}^2 \int_{y=0}^{2-x}[2 x z]_{z=0}^{4-2 x-2 y} d y d x\)

= \(\int_{x=0}^2 \int_{y=0}^{2-x} 2 x(4-2 x-2 y) d y d x=\int_{x=0}^2 \int_{y=0}^{2-x}\left(8 x-4 x^2-4 x y\right) d y d x\)

= \(\int_{x=0}^2\left[8 x y-4 x^2 y-2 x y^2\right]_{y=0}^{2-x} d x=\int_0^2\left[8 x(2-x)-4 x^2(2-x)-2 x(2-x)^2\right] d x\)

= \(\int_0^2\left(2 x^3-8 x^2+8 x\right) d x=\left[\frac{2 x^4}{4}-\frac{8 x^3}{3}+\frac{8 x^2}{2}\right]_0^2=\frac{1}{2} \cdot 2^4-\frac{8}{3} \cdot 2^3+4 \cdot 2^2=\frac{8}{3}\)

Integral Transformations Gauss’s Divergence Theorem

Let S be a closed surface enclosed in a volume V . If F is a continuously differentiable vector point function, then \(\int_V d i v \mathbf{F} d \mathbf{V}=\int_S \mathbf{F} \cdot \mathbf{N} d S\)

Where N is the outward drawn unit normal vector at any point of S.

Green’s identities And Gauss’s Divergence Relation

Cartesian Form

Let F =F₁i+F₂ j+F₃ k and N=i cos α+j cos β+k cos γ

where \(\cos \alpha, \cos \beta, \cos \gamma\) are the direction cosines of N. \(\mathbf{F} . \mathbf{N}=\mathbf{F}_1 \cos \alpha+\mathbf{F}_2 \cos \beta+\mathbf{F}_3 \cos \gamma\)

Also div \(\mathbf{F}=\nabla . \mathbf{F}=\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\)

Hence the divergence theorem can be written as \(\iiint\left(\frac{\partial \mathbf{F}_1}{\partial x}+\frac{\partial \mathbf{F}_2}{\partial y}+\frac{\partial \mathbf{F}_3}{\partial z}\right) d x d y d z\)

= \(\int_S\left(\mathrm{~F}_1 \cos \alpha+\mathrm{F}_2 \cos \beta+\mathrm{F}_3 \cos \gamma\right) d S\)

= \(\iint_S\left(\mathbf{F}_1 d y d z+\mathbf{F}_2 d z d x+\mathbf{F}_3 d x d y\right)\)

Proof: Let s be a closed surface. Let us choose the coordinate axes so that any line parallel to the coordinate axes cuts S in at most two points.  Let R be the projection of S on xy-plane. S and S are the lower and upper parts of S.

z=f(x,y) and z=g(x,y) be the  equations of S₁ and S₂. The relation can be put in the form f(x,y) ≤ z ≤ g(x,y).

∴  \(\int_V \frac{\partial \mathbf{F}_3}{\partial z} d \mathbf{V}\)\(=\iiint_V \frac{\partial \mathbf{F}_3}{\partial z}\) dx dy =\(=\iint_R\left[\frac{\partial \mathbf{F}_3}{\partial z} d z\right]_{z=f(x, y)}^{z=g(x, y)}\) dx dy

= \(\iint_R\left[\mathbf{F}_3(x, y, z)\right]_f^g d x d y=\iint_R\left[\mathbf{F}_3(x, y, g)-\mathbf{F}_3(x, y, f)\right] d x d y\)

= \(\iint_R \mathrm{~F}_3(x, y, g) d x d y-\mathrm{F}_3(x, y, f) d x d y\) …….. (1)

For the upper part \(\mathbf{S}_2 \quad d x d y=d S \cdot \cos \gamma=\mathbf{N} \cdot \mathbf{k} d S\)

Since the normal to \(\mathbf{S}_2\) makes an acute \(\gamma\) with \(\mathbf{k}\).

∴ \(\iint_R \mathbf{F}_3(x, y, g) d x d y=\int_{S_1} \mathbf{F}_{\mathbf{3}} \mathbf{N} \cdot \mathbf{k} d S\)

For the lower portion \(\mathbf{S}_1 \quad d x d y=-\cos \gamma d S=-\mathbf{N} \cdot \mathbf{k} d S\)

Since the normal to \(\mathbf{S}_1\) makes an obtuse angle \(\gamma\) with \(\mathbf{k}\)

∴ \(\iint_R \mathbf{F}_3(x, y, f) d x d y=-\int_{S_3} \mathbf{F}_3 \mathbf{N} \cdot \mathbf{k} d S\)

Hence from (1)

∴ \(\int_V \frac{\partial \mathbf{F}_3}{\partial z} d \mathbf{V}=\int_{S_2} \mathbf{F}_3 \mathbf{k} \cdot \mathbf{N} d S+\int_{S_1} \mathbf{F}_3 \mathbf{k} \cdot \mathbf{N} d S=\int_S \mathbf{F}_3 \mathbf{k} \cdot \mathbf{N} d S\)

The theorem can be extended to surfaces which are such that lines parallel to coordinate axes meet them in more than two points. In such a case, we subdivided the region bounded by S into subregions whose surfaces satisfy this condition. Applying the same procedure we can prove the Theorem.

Green’s Identities Explained With Examples

Integral Transformations Deduction From Gauss’s Theorem

1. Prove that\(\int_S\)N × F ds=\(\int_V\) ∇×F dv

Proof: Let f=a × F, where a is a constant vector.

Applying Gauss’s Theorem on f we have \(\int_S\)f.Nds= \(\int_V\)∇f dv ⇒ \(\int_S\)(a×f).Nds=\(\int_V\)∇.(a×f) dv

⇒ \(\int \mathbf{a} \cdot(\mathbf{F} \times \mathbf{N}) d \mathbf{S}=-\int \nabla \cdot(\mathbf{F} \times \mathbf{a}) d \mathbf{V}=-\int(\nabla \times \mathbf{F}) \cdot \mathbf{a} d \mathbf{V}\)

because \(\mathbf{a}\) is constant

⇒ \({\mathbf{a} \cdot \int_S}(\mathbf{N} \times \mathbf{F}) d \mathbf{S}=\mathbf{a} \cdot \int_S \nabla \times \mathbf{F} d \mathbf{V}\)

⇒ \(\mathbf{a} \cdot\left[\int_S(\mathbf{N} \times \mathbf{F}) d \mathbf{S}-\int_V \nabla \times \mathbf{F} d \mathbf{V}\right]=0\)

Since a is a constant vector

⇒ \(\int_S(\mathbf{N} \times \mathbf{F}) d \mathbf{S}-\int_V \nabla \times \mathbf{F} d \mathbf{V}=0 \Rightarrow \int(\mathbf{N} \times \mathbf{F}) d \mathbf{S}=\int \nabla \times \mathbf{F} d \mathbf{V}\)

2. Prove that \(\int_S\)NΦ ds=\(\int_V\)∇Φ dv

Proof:

Applying Gauss theorem to \(\mathbf{a} \phi\) we have

⇒ \(\int_S(\mathbf{a} \phi) \cdot \mathbf{N} d \mathbf{s}=\int_V \text{div}(\mathbf{a} \phi) d \mathbf{V}=\int_V \nabla \cdot(\mathbf{a} \phi) d \mathbf{V}\)

a. \(\int_S(\phi \mathbf{N}) \cdot \mathbf{N} d \mathbf{s}=\mathbf{a} \cdot \int_V(\nabla \phi) d \mathbf{V} \Rightarrow \mathbf{a} \cdot\left[\int_S \phi \mathbf{N} d \mathbf{S}-\int_V \nabla \phi d \mathbf{V}\right]=0\)

Since a is a constant vector.

⇒ \(\int_S \phi \mathbf{N} d \mathbf{S}-\int_V \nabla \phi d \mathbf{V}=0\)

∴ \(\int_S \mathbf{N} \phi d \mathbf{S}-\int_V \nabla \phi d \mathbf{V}=0\)

Note. The result proved above can be rewritten as follows.

1.\(\int_S\)Φ N.F dS = ∫∇.F dV

2.\(\int_S\)N× F dS=\(\int_V\)∇× F dV

3.∫NΦ dS=∫∇Φ dv

Here N is written before the function in L.H.S ∇ displaces N in R.H.S

Relationship Between Gauss’S Divergence Theorem And Green’S Identities

Integral Transformations  Green Identities

If f and g are two continuous and differentiable scalar point functions over the region V enclosed by the surfaces S. then

1. \(\int_V\)[f∇² g+∇ f.∇ g]dV=\(\int_S\)(f.∇ g).N dS
2. \(\int_V\)[f∇² g-g∇² f]dV=\(\int_S\)(f∇g-g∇f).N S

Proof: By Gauss’s divergence theorem we have \(\int_V\)∇. F dV=\(\int_S\)F.N dS

Now substituting \(\overline{\mathrm{F}}\)=f ∇g, we get

∇.F ∇(f∇g)=f(∇.∇g)∇f+∇g

∴ Divergence theorem gives \(\int_V\)[f∇²g+∇f.g]dV=\(\int_S\)(f∇g).N dS      ……………….(1)

This is called Green’s first identity or theorem.

2. Interchanging f and g in (1), we get

∫[g∇²f+∇g.∇f]dV=(g∇f).N dS     ………………(2)

Subtracting (2) from(1) we, get

∫(f∇²g-g∇²f)dV=∫f∇g-g∇f). N dS

This is called Green’s second identity or Green’s theorem in symmetrical form.

Subrings, Ideals, Quotient Rings & Euclidean Rings Quotient Rings Or Factor Rings

The concept of a quotient ring is analogous to that of quotient groups. If U is an ideal of a ring (R,+,•) then (U,+) is a normal subgroup of the commutative group (R,+).

From group theory, we know that the set R/U.= {x+U =U+ x | x ∈ R} of all cosets of U In R is a group with respect to the addition of two cosets defined by (a+U) + (b+U) = (a+b)+U for a+U,b +U ∈(R/U).

We know further that these costs are disjoint.

As addition operation is commutative left coset a+llis equal to right coset u + a In order to impose ring structure in R/U we can define multiplication of two of cosets as (a+U) (b+U) = ab +U for a+U,b +U ∈ R /U

Subrings, Ideals, Quotient Rings, And Euclidean Rings Theorems

Theorem.1 If U is an ideal of a ring R then the set R/U ={x+U|x ∈R} is a ring with respect to the induced operations of addition (+ ) and multiplication ( • ) of cosets defined : (a+ u)+(b+u)=(a+b)+U and (a+U). (b+U)=ab+U for a+Ub+u R/U.

Proof. Since (R,+) is a commutative group, the quotient group (R/U,+) is also commutative, In order to show that (R,+,•) is a ring we must show that

(1) multiplication of cosets is well defined,
(2) multiplication is associative and (3) distributive laws hold.

(1) Let \(a+U=a_1+U\) and \(b+U=b_1+U \text {. }\)

Then \(a=a_1+u_1\) and \(b=b_1+u_2\) for \(u_1, u_2 \in U \text {. }\)

ab = \(\left(a_1+u_1\right)\left(b_1+u_2\right)=a_1 b_1+a_1 u_2+u_1 b_1+u_1 u_2\)

Since U is an ideal, \(a_1 u_2, u_1 b_1, u_1 u_2 \in U\)

∴ \(a b-a_1 b_1 \in U\) and hence \(a b+U=a_1 b_1+U \Rightarrow(a+U) \cdot(b+U)=\left(a_1+U\right) \cdot\left(b_1+U\right)\)

Therefore multiplication of cosets is well defined.

Let a+U,b+U,c+U ∈R/U

(2) [(a+U).(b+U)] .(c+U) = (ab+U).(c+U) = (ab) c+U = a'(bc) +U (v a,b,c∈R) = (a +U). (bc+U) = (a+U) . [(b+U).(c+U)]

(3) (a+U). [(b+U)+(c+U)] = (a+U).[(b+ c)+U] = a(b+ c)+U = (ab+ ac)+U (v a,b,c∈R)

(ab+U)(ac+U) = (a+U) .(b+U)+(a+U). (c+U)

Similarly we can prove that [(b+U)+ (c+ U)].(a+U) = (b+U).(a+U)+(c+U).(a+U) Hence (R/U,+,•) is a ring

Examples Of Factor Rings In Abstract Algebra

Definition Of Addition And Multiplication Cosets

Definition. Let R be a ring and U be an ideal of R. Then the set R/U = {x+U\x∈R] with respect to induced operations of addition and multiplication of cosets defined by (a+U)+(b+U)-(a+b) +U;(a+U). (b+U) = ab +Ufor a+U,b+U eR/U is a ring.

This ring (R/U,+,•)is called the quotient ring or factor ring, or residue class ring of R modulo U.

Note.

1. It is convenient, sometimes, to denote coset a+U in R/U by the symbol a or.[a]. Then we write the sum and product of two cosets as [a]+[b] =[a+b] and [a].[b] = [ab].
2. o+U = U is the zero element in the ring R/U.
3. Every ring R has two improper ideals, namely, the trivial ideal {0} and the ideal R. The quotient ring of the ideal {0} is R/{0} or R /(0) = {x+ (0)| x∈ R}The quotient ring of the ideal R is R / R or Rt(l) = {x+(l) | x ∈ R}
4. (a+U)+ (b+U) = (a+b)+U-,(a+U)(b+U) = ab+U
5. (a+U)² =(a+U)(a+U) = a²+U
6. a+U = b+U ⇔(a-b)∈U.
7. a +U =U ⇔ a ∈ U

Theorem. 2. If R/U is the quotient ring prove that (1) R/U is commutative if R is commutative and (2) R/u has a unity element if R has a unity element.

Proof. (2) R is commutative => ab = ba∀ a,b ∈ R.

Let a+U,b+U ∈ R/U. (a+U) (b+U) = ab+U = ba+U =(b+U).(a+U)

∴ R/U is commutative.

(2) R has unity element => there exists 1 ∈ R so that a1 = 1a- a∀ a∈R.

Let a+U ∈ R/U.  For 1∈R we have 1+U ∈R/U

We now prove that 1 +U is the unity element.

(a+U)(l+U) = a1+U = a+U and (X+U)(a+U)*=la+U = a+U V a+U eR/U

1+U is the unity element in R/U.

Note. In the quotient ring R/U, the unity element = 1 +U.

Rings Examples Of Addition And Multiplication Cosets

Example. 1 Consider Z6 = {0,1, 2,3,4,5}, the ring of integers modulo 6.

U= {0,3} is an ideal of Z6. The costs of Uin R are as follows :

0+U = {0 + 0, 0+ 3} = {0,3};l+t/ = {1 + 0,1+ 3} = {1,4}

2+U = {2 +02 + 3} = {2,5};3+t/ = {3+0, 3+ 3} = {3,0} =0+ 17

4+U = {4+ 0, 4+ 3} = {4, 1} – 1 +U and 5+U {5 + 0, 5+3} = {5, 2} = 2+U

(Z6/U) = {0+U,l+U,2+U} is the quotientring.

Note. We observe that two cosets are identical or disjoint and the union of all cosets = Z6.

Quotient Rings And Their Properties Explained

Example 2. For the ring Z of all integers, we know that nZ = {nx| x∈ Z} for any n∈ Z is an additive subgroup of Z.

Let m ∈ nZ and r ∈ Z  Then m = na where a∈ Z.

mr = (na)r = n (ar) and rm =r (na) = n (ar)

so that mr = rm = n (ar) = nb∈nZ where b = ar∈Z Thus nZ is an ideal of Z.

The set of all cosets of nZ in Z, namely, (Z/nZ) = {x+nZ |x∈ Z} forms a ring under the induced operations of addition and multiplication.

Euclidean Rings

Definition. An integral domain R is said to be a Euclidean ring or Euclidean domain if for every a(≠0)∈ R there is defined a non-negative integer d(a) such that

(1)for all a,b∈ R, a≠ 0,b≠ 0;d (a) < d (ab) and

(2)for any a,b∈R,b≠ 0 there exist q,r∈R such that a = bq +r where either r = 0 or d (r) <d(b) .

Note

1. For any a(≠ 0)∈R,d (a)> 0.
2. For the zero element 0 of R,d{0) is not defined. However, some authors defined d(0) = 0, integer.
3. The property (2) in the above definition is called the division algorithm.
4. From the above definition, we note that d: R- {0} -> Z is a mapping such that

(1) d (a) > 0 ∀ R— {0} .
(2) d (a) < d (a,b)∀1 a,b∈R- {0} and
(3) there exist q,r∈R so that a-bq+r where either r = 0 or d (r) < d (b) for any a b∈R and b≠0.

Subrings, Ideals, Quotient Rings, And Euclidean Rings Theorems

Theorem 1. Every field is a Euclidean ring

Proof. Let F be a field and F be the set of all non-zero elements of F. Since F is a field, F is an integral domain

Define the mapping d:FZ by d(a) = 0 (zero integer) for all a ϵ F

d (a) ≥0 ∀ a ϵF

Let a bϵF.

Then a,b, and ab are non-zero elements of F

d (a) = 0 and d (ab) = 0 => d (a) <,d (ab)

Let aϵF and bϵF.

Now a = a 1 where 1 is the unity element of F.

=a(b^-1b) = (ab^-1)b  ( b^-1b=1)

= (ab^-1]) b+ 0 where ‘O’ is the zero element of the field F.

a = qb +r where q = ab^-1,r = 0 1

Hence, for a ϵ F,b ϵ F there exists q,r ϵ F so that a = qb+r where r= 0. F is an Euclidean ring.

Note. We can prove the above theorem by defining d: F -> Z by d(a) = 1 (integer) ∀ a ϵ F

Theorems On Subrings And Their Applications

Theorem 2. Every Euclidean ring is a principal ideal ring (or) Every ideal of an Euclidean ring is a principal ideal

Proof. Let R be an Euclidean ring.

Let U = {0} where ‘O’ is the zero element of R. Then U = {0} is the ideal generated by QϵR.

U is a principal ideal of R.

Let U be an ideal of R.

Let u≠ {0}. there exists xϵU and x≠0 so that the set {d (x) | x≠ 0} is a non-empty set of nonnegative integers.

By well ordering principle there exists b≠0 ϵ U so that d(b)<d (x) where x≠0ϵU.

We now prove that U = (b). Let ‘a’ be any element of U.

By division algorithm, there exists q,rϵR so that a -bq +r where r = 0 or d (r) < d (b). bϵU,qϵR, and U is an ideal => bqϵU. aϵU,bq ϵU => a-bq =r ϵU

If r≠ 0 then d (r) < d (b) so that we have a contradiction as d (b) < d (x) V x≠ 0 ϵU

r = 0 and hence a=b q.

U- {bq| qϵR} = (b) is the principal ideal generated by b(≠ 0) ϵU. Hence every ideal U of R is a principal ideal.

R is a principal ideal ring.

Note 1. If U is an ideal of a Euclidean ring R then U is a principal ideal of R so that U = (b) = {bq |q ϵF}

For,’ the ring R = \(\left\{a+b\left(\frac{1+\sqrt{19 i}}{2}\right): a, b \in Z\right\}\) of complex numbers is a principal ideal ring but not Euclidean

Subrings, Ideals, Quotient Rings & Euclidean Rings Principal Ideal

If a commutative ring with unity from Theorem (3) Art! 2.2 we observed that for a given a ∈ R the set {ra | r ∈ #} is an ideal in R that contains the element ‘a’.

Definition. Let R be a commutative ring with unity and a ∈ R The ideal {ra |r ∈ R) of all multiples of ‘a’ is called the principal ideal generated by ‘a’ and is denoted by (a) or (a).

An ideal Uofthe ring R is a principal ideal =>U = (ci) = {ra ) r ∈ It} for some a∈R

example. 1 The null ideal or trivial ideal {0}’ofaring# is the principal ideal generated by the zero element of R. That is null ideal = (0).

example. 2. The unit ideal or improper ideal R of a ring R is the principal ideal generated by the unity element T of the ring R. That is R =(l).

example.3. Z is a commutative ring with a unity element. The principal ideal generated by 2 ∈ Z = (2) = [2n| n ∈ Z} = the set of all even integers.

example. 4. A field has only null ideal = (0) and unit ideal =F =(l) which are principal ideals.

Definition. (Principal ideal ring). A commutative ring R with unity is a principal ‘ ideal ring if every ideal in R is a principal ideal

D is a principal ideal domain => every ideal U in D is in the form U (a) for some a ∈D.

Principal Ideal Theorem In Euclidean Rings Explained

Theorem. 1. A field is a principal ideal ring

Proof. A field F has only two ideals, namely, U = (0) and U = F = {l).

But U = (0) and U = (l) are principal ideals. the field F is a principal ideal ring.

Theorem. 2. ring of integers the principal ideal ring, (or) Every ideal Z is of Z is a principal ideal.

Proof. Let U be the ideal of Z and U = {0}. Then U is generated by the zero element.

.’. U = (0) is a principal ideal. Let U be an ideal of Z and U≠ (0).

there exists a ∈U so that a ≠  0,

Since u ⊂ Z, one of a, -a must be a positive integer.

∴ the set of positive integers U+ in U is non-empty. by the well-ordering principle, U+ has the least member, say, b.

We now prove that U = {b) = the principal ideal generated by ‘b’.

Let x∈U. Since x,b are integers and b≠0 there exist q,r ∈ Z

such that x = bq+r; 0 <,r<b (Division algorithm).

b ∈ U,q ∈ Z and u is an ideal => bq ∈ U. x∈U,bq ∈U => x-bq =r ∈U .

Now r ∈U 0 <r < b and b is the least member in U+ =>r = 0.

x-bq =r=>x-bq = 0 => x = bq .

Hence x∈U=>x=bq for q∈Z =>U = (b).

∴  every ideal U of Z is a principal ideal. Hence Z  is a principal ideal ring.

Examples Of Principal Ideal Theorem In Subrings And Quotient Rings

Note. 1. Principal ideal rings that are also integral domains, such as rings of integers Z  are called principal ideal domains

2. If Z is the ring of integers then the principal ideal generated by a∈ Z is ] {aq \q∈Z} = (a).

Subrings, Ideals, Quotient Rings & Euclidean Rings Ideas

The concept of an ideal of a ring is analogous to that of a normal subgroup of a group. Some of the subrings which we call ideals play a very important role as the normal subgroups
in group theory.

Definition. (Ideal). Let. (R, +,• ) be a ring. A non-empty subset U of R is called a two-sided ideal or ideal if (1) a,b ∈ U=>a-b∈U and (2) a ∈ U and r ∈ R=> ar,r a∈ U.

(Ideal). A subring U of a ring R is called a (two-sided) ideal of R if, for every r ∈ R and every a ∈ U, both ra and ar are in U.

Note

1. A subring u of the ring R satisfying rU⊂ U and Ur⊂U for all r ∈ R is an ideal.
2. The (2) condition of ideal is stronger than the (2) condition of a subring.
3. The condition (1), namely, a,b∈ U => a -b∈U is called module property.
4. If U is an ideal of the ring R,+, • ) then (U, +) is a normal subgroup of the commutative group (R, +). Hence zero element in R is zero element in U.

Definition. A non-empty subset U of a ring R is called a right ideal if.

(1) a,b∈U=>a-b ∈U and(2)a ∈U,r ∈R=>ar ∈U . A non-empty subset U of a ring R is called a left ideal if (1) a,b ∈U a-b ell and a ∈U,r∈R=>ra ∈U .

Note.

1. An ideal is both a left and a right ideal.
2. For commutative rings left ideals coincide with right ideals.

Theorem 1. If U is an ideal of a ring R with unity element and 1∈ U then U  =R.

Proof: by the definition ideal  U⊂R.

Also x ∈R=>x.1 eR =>x.1∈U for x∈R, 1 ∈u (Def. of ideal) =>x ∈U .

R⊂U and hence U ⊂ R

Theorems On Ideals In Subrings And Quotient Rings

Theorem 2. A field has no proper non-trivial ideals, (or) The ideals of a field f are only {0} and F itself

Proof. Let U be an ideal of F so that U ≠ {0}

We now prove that U = F

By the definition of ideal, U⊂ F …… (1)

Let a∈ U and a≠ 0.

For a (≠ 0) ∈F there exists a -1  ∈  F so that aa -1 = 1

x ∈F => x 1 ∈ F=>x . 1 ∈U for xeF, 1∈U

=> x∈U .-. F ∈U … (2)

From (1) and (2): U = F Hence ideals of F are either {o} orF.

Theorem 3. If R is a commutative ring and a ∈ R then Ra = {ra\r| ∈ R \ is an ideal of R.

Proof. For 0 ∈ R, 0a = 0∈Ra. Ra≠ Φand R⊂R.

Let x,y∈Ra. Then x =r 1 a, y = r2a where r1,r2 ∈ R  :. x,y ∈ Ra=>x-y ∈ R

x. y = r1a-r2a=(r1-r2)a= ra where r =r1~r2 ∈ R.

Let x ∈ Ra and r ∈R.

x.r = (r1a)r ( :.x = r1a Where r∈R) =r1(ar) = r1(ra) (By R5 and R is commutative )

=(r1r) a-r’a where r’ = r1r ∈ R .

Since R is commutative, x.r=r.x.

∴ x∈Ra, r ∈R=>xr = rx ∈Ra  ……. (2)

Hence from (1) and (2): Ra is an ideal of R.

Note. 1. If R is a commutative ring and a e R then aR = { ar \ r e R } is an ideal of R.

2. If R is a r in and a ∈ R then Ra is a left ideal and aR is a right ideal.

Theorem 4. A commutative ring R with unity element is a field if R have no proper ideals.

Proof. Since the Ring R has no proper nontrivial ideals, the ideals of R are {o} and R only.

To prove that R is a field we have to show that every a (≠ 0) ∈ R has a multiplicative inverse. We know that aR = {ar|r ∈ R} is an ideal of R.

Since a ≠ 0 .aR ≠ {0} and hence aR =R (By hypothesis)

1 ∈ R => 1 ∈ a R =>1 = ab for some b≠ 0 ∈ R . Since R is conimutatiye,1= ab= ba .

a ≠ 0 ∈ R has a multiplicative inverse b∈ R . Hence R is a field.

Note. If R is a ring with a unity element and R has no proper nontrivial ideals then R is a division ring.

Examples Of Ideals In Quotient Rings And Euclidean Domains

Theorem. 5. The intersection of two ideals of a ring R is an ideal of R.

Proof. Let U1, and U2 be two ideals of the ring R.

If 0 ∈ R is the zero element, then 0∈ U1 and 0 ∈ U2 .

:. 0∈U1∩U2 and hence U1∩ U2 ≠φ

Let a,b ∈ U1 ∩ U2 and r ∈ R. Then a b∈U and a,b ∈U2.

a,b ∈ U1,r ∈R and U1 is an ideal =>a-b ∈U1 and ar, ra ∈ U1……………………….1

a,b ∈U2,r ∈R and U2 is an ideal=> a-b ∈U2 and ar, ra ∈U2………………………….2

From (1) and (2): a-b ∈ U1 ∩ U2 and ar, ra ∈ U1∩U2

Hence U1∩ U2 is an ideal of R.

Remark: The union of two ideals of a ring R need not be an ideal of R

For the ring Z of integers, A = { 2n | n∈ Z) and B = {3n\n ∈ Z } are two ideals.

But, for 2,3 ∈ A∪B, 3-2 = 1 ≠ A∪B.   A∪B is not an ideal of Z

Understanding The Properties Of Ideals In Algebraic Structures

Theorem 6. If U₁, and U₂ are two ideals of a ring R then U₁∪U₂is ideal of R if and only if U₁⊂ U₂ or U₂⊂U₁

Proof. Let U₁ ∪ U₂ be an ideal of R. We now prove that U₁⊂ U₂ or U₂⊂U₁

If possible, suppose that U₁⊄ U₂ or U₂⊄U₁

Since U₁⊄ U₂ there exists an element a ∈ U₁ and a ∈ U₂.

Since U₂⊄U₁ there exists an element b ∈ U₂ and ∈ U₁.

a ∈ U₁ and b∈ U₂  => a, b ∈ U₁⊂ U₂

a,b ∈ U₁⊂ U₂ and U₁⊂ U₂ is an ideal =>a-b ∈ U₁ ∪ U₂

=> a-b ∈ U₁  or a-b ∈ U₂

But a-b ∈ U₁=>a- {a-b) = b ∈U₁ …………………..1

a-b ∈ U₂ => b+(a-b)=a ∈ U₂    ………………………2

Both (1) and (2) contradict a ∉U₂ and b∉U₁

Our supposition is wrong.

Hence U₁⊂ U₂ or U₂⊂U₁

Conversely, let U₁⊂ U₂ or U₂⊂U₁

Then U₁ ∪ U₂  = U2 or U₁and is an ideal.

Note. If S₁, and S₂ are two subrings of a ring prove that S₁ u S₂ is also a subring iff either S_{1 ⊆} S₂ or S₂ ⊆S₁