Q. 2 - Calculus

Question

2. Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.
(a) An iterated integral that represents the area of a circle with radius $R$ express with polar coordinates.
(b) An iterated integral using polar coordinates that represents the volume of a sphere with radius $R$.
(c) An iterated integral in rectangular coordinates that would be easier to evaluate by using polar coordinates.

Step-by-Step Solution

Answer

(a) The required area of the circular ground is $A=\pi R^2$

(b) The volume of a sphere is $V=\frac{4}{3}\pi a^3$

(c) ${\int }_0^2{\int }_0^{\sqrt{2x-x^2}}\left(x^2+y^2\right)dydx=\frac{3\pi }{4}$ is true

Part(a) Step 1 Given Information

The objective of this problem is to construct an example of area of a circle with radius $R$.

Part (a) Step 2 Calculation and Diagram

Find the area of a lawn grazed out by a cow. The cow is tied with a rope of length $R$ meter. Consider a circle of radius $R$

Area of a sector of a circle of radius $R$ enclosed by $\theta =\alpha$  and $\theta =\beta$ can be expressed as $\frac{1}{2}{\int }_{\theta -\alpha }^{\theta -\beta }{r}^{2}d\theta$
Area of a quarter of a circle can be expressed as a sum of area of four sectors.
Area of a circle $=$ Sum of area of four quarters
Area of a quarter of a circle $=\frac{1}{2}{\int }_0^{\pi /2}{r}^{2}d\theta$
Therefore, area of a circular ground
$A=4\times \frac{1}{2}{\int }_0^{\pi /2}{r}^{2}d\theta$

$A=4\times \frac{1}{2}{\int }_0^{\pi /2}{R}^{2}d\theta [r=R]$

$A=2R^2{\int }_0^{\pi /2}d\theta$

Integrate with respect to $\theta$.
$A=2R^2[\theta {]}_0^{*/2}$
Put the limits
$A=2R^2\left[\frac{\pi }{2}-0\right]$

$A=\pi R^2$
Thus, the area of a circular ground is
$A=\pi R^2$

Part (b) Step 1 Given Information

Take an example of sphere of radius } a \text { and center at origin.

Part (b) Step 2 Calculation

The equation of sphere is
$x^2+y^2+z^2=a^2$
The sphere is symmetrical about $x-y$ plane. Therefore, its volume can be computed as the upper half of sphere multiplied by 2
$V=2{\int }_{x-a}^{x-a}{\int }_{y-0}^{\sqrt{a^2-x^2}}zdxdy$

$V=2{\int }_{x-a}^{x-a}{\int }_{y=0}^{\sqrt{a^2-x^2}}\sqrt{a^2-\left(x^2+y^2\right)}dxdy\left[x^2+y^2+z^2=a^2\right]$
For the polar form
Substitute $x=r\cos \theta ,y=r\sin \theta$ and $dxdy=rdrd\theta$
Where, $0\le r\le a$ and $-\pi \le \theta \le \pi$
$V=2{\int }_{\theta =-\pi }^{\theta -s}{\int }_{r=0}^{r-a}\sqrt{a^2-\left(r^2{\cos }^2\theta +r^2{\sin }^2\theta \right)}rdrd\theta$

$V=2{\int }_{x=-a}^{x-a}{\int }_{y=0}^{\sqrt{a^2-x^2}}\sqrt{a^2-\left(x^2+y^2\right)}dxdy\left[x^2+y^2+z^2=a^2\right]$
$V=2{\int }_{\theta -n}^{\theta -\pi }{\int }_{r=0}^{r=a}\sqrt{a^2-r^2}rdrd\theta \left[{\sin }^2\theta +{\cos }^2\theta =1\right]$
Integrate with respect to $r$.

$V=2{\int }_{\theta -\pi }^{\theta -\pi }{\left[-\frac{1}{3}{\left(a^2-r^2\right)}^{3/2}\right]}_0^{a}d\theta$
$V=2{\int }_{\theta -\pi }^{\theta -\pi }\left[\frac{1}{3}{\left(a^2\right)}^{3/2}\right]d\theta$

$V=\frac{2}{3}{a}^3{\int }_{\theta -\pi }^{\theta -\pi }d\theta$

Integrate with respect to $\theta$.
$V=\frac{2}{3}{a}^3[\theta {]}_{-z}^z$

$V=\frac{2}{3}{a}^3[\pi -(-\pi \left)\right]$

$V=\frac{4}{3}\pi a^3$

Thus, the volume of a sphere is
$V=\frac{4}{3}\pi a^3$

Part (c) Step 1 Given Information

Consider an example of double integration.
$I={\int }_0^2{\int }_0^{\sqrt{2x-x^2}}\left(x^2+y^2\right)dydx$

Part (c) Step 2 Calculation

$\backslash begin\left\{aligned\right\} \&I=\backslash int\_\left\{0\right\}^\{\backslash pi / 2\} \backslash int\_\left\{0\right\}^\{2 \backslash \cos \backslash theta\}\backslash left(r^\{2\} \backslash \cos ^\{2\} \backslash theta+r^\{2\} \backslash \sin ^\{2\} \backslash theta\backslash right) r d r d \backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{x / 2\} \backslash int\_\left\{0\right\}^\{2 \backslash \cos \backslash theta\} r^\left\{3\right\} d r d \backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left[\backslash frac\{r^\left\{4\right\}\left\}\right\{4\}\backslash right]\_\left\{0\right\}^\{2 \backslash operatorname\{ees\} \backslash theta\} d \backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left[\backslash frac\{16\left\}\right\{4\} \backslash \cos ^\{4\} \backslash theta-0\backslash right] d \backslash theta\backslash \backslash \&I=4 \backslash int\_\left\{0\right\}^\{\backslash pi / 2\} \backslash \cos ^\left\{4\right\} \backslash theta d \backslash theta\backslash \backslash \&I=4 \backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left(\backslash \cos ^\{2\} \backslash theta\backslash right)^\left\{2\right\} d \backslash theta\backslash \backslash \&I=4 \backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left(\backslash frac\{1+\backslash \cos 2 \backslash theta\left\}\right\{2\}\backslash right)^\left\{2\right\} d \backslash theta\backslash \backslash \&I=4 \backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left(\backslash frac\{1+2 \backslash \cos 2 \backslash theta+\backslash \cos ^\left\{2\right\} 2 \backslash theta\left\}\right\{4\}\backslash right) d \backslash theta\backslash \backslash \&I=\backslash int\_\left\{0\right\}^\{\backslash pi / 2\}\backslash left\backslash \{1+2 \backslash \cos 2 \backslash theta+\backslash frac\{1\left\}\right\{2\left\}\right(1+\backslash \cos 4 \backslash theta)\backslash right\backslash \} d \backslash theta\backslash \backslash \&I=\backslash left[\backslash theta+\backslash \sin 2 \backslash theta+\backslash frac\{1\left\}\right\{2\}\backslash left(\backslash theta+\backslash frac\left\{1\right\}\left\{4\right\} \backslash \sin 4 \backslash theta\backslash right)\backslash right]\_\left\{0\right\}^\{\backslash pi / 2\}\backslash \backslash \&I=\backslash left[\backslash frac\{\backslash pi\left\}\right\{2\}+\backslash \sin (\backslash pi)+\backslash frac\{1\left\}\right\{2\}\backslash left(\backslash frac\{\backslash pi\}\left\{2\right\}+\backslash frac\left\{1\right\}\left\{4\right\} \backslash \sin 2 \backslash pi\backslash right)\backslash right] \backslash end\left\{aligned\right\}$

Here upper limit of $y$ is

$y=\sqrt{2x-x^2}$ 

$y^2=2x-x^2\text{ [Square both sides] }$
$\Rightarrow x^2+y^2+2x=0\dots \dots \text{ (1) }$

Equation (1) represents a circle with center (1,0)  and unit radius.

Lower limit of $y$ is 0 .
Region of integration is the upper half of circle.
Substitute $x=r\cos \theta ,y=r\sin \theta$ in equation (1)
$r^2-2r\cos \theta =0\Rightarrow r(r-2\cos \theta )=0$

$r=0,r=2\cos \theta$

$\theta =0,\theta =\frac{\pi }{2}$
Therefore,

$I={\int }_0^{\pi /2}{\int }_0^{2\cos \theta }\left(r^2{\cos }^2\theta +r^2{\sin }^2\theta \right)rdrd\theta$
$I={\int }_0^{\pi /2}{\int }_0^{2\cos \theta }{r}^{3}drd\theta$
$I={\int }_0^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^{2\cos \theta }d\theta$
$I={\int }_0^{\pi /2}\left[\frac{16}{4}{\cos }^4\theta -0\right]d\theta$

$I=4{\int }_0^{\pi /2}{\cos }^4\theta d\theta$

$I=4{\int }_0^{\pi /2}{\left({\cos }^2\theta \right)}^{2}d\theta$

$I=4{\int }_0^{\pi /2}{\left(\frac{1+\cos 2\theta }{2}\right)}^{2}d\theta$

$I={\left[{\int }_0^{\pi /2}\left(\sin \left(\pi \right)+\frac{1}{2}\left(\frac{\pi }{2}+\frac{1}{4}\sin 2\pi \right)\right]\sin 2\theta +\frac{1}{2}\left(\theta +\frac{1}{4}\sin 4\theta \right)\right]}_0^{\pi /2}$

$I=\left[\frac{1+2\cos 2\theta +{\cos }^{2}2\theta }{4}d\theta \right.$

$I=\left[\frac{\pi }{2}+\sin \left(\pi \right)+\frac{1}{2}\left(\frac{\pi }{2}+\frac{1}{4}\sin 2\pi \right)\right]$

$I=\frac{3\pi }{4}$

${\int }_0^2{\int }_0^{\sqrt{2x-x^2}}\left(x^2+y^2\right)dydx=\frac{3\pi }{4}$

True