Q. 2 - Calculus

Question

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 circular ground of area is $A={\mathrm{\pi r}}^2$

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

(c) The double integral over a polar rectangular region can be stated as an iterated integral in polar coordinates, much like in Double Integrals over Rectangular Regions.

Part (a) step 1: Given information

The circular ground of area is  $\mathrm{A}={\mathrm{\pi r}}^2$

Part (a) Step 2: Calculation of the area of a circle

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

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

Consider the radius R of a circle.

Area of a sector of a circle of radius   can be expressed as  

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

a quarter of a circle's area = $=\frac{1}{2}{\int }_0^{\mathrm{\pi }/2} {\mathrm{r}}^2\mathrm{d\theta }$

As a result, the surface area of a circular ground 

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

Integrate with respect to $\mathrm{\theta },$

$\mathrm{A}=2{\mathrm{R}}^2[\mathrm{\theta }{]}_0^{\mathrm{\pi }/2}$

Put the limits 

$$\begin{gathered} \mathrm{A}=2{\mathrm{R}}^2\left[\frac{\mathrm{\pi }}{2}-0\right] \\ \mathrm{A}={\mathrm{\pi R}}^2 \end{gathered}$$

As result, the surface area of a circular ground is $\mathrm{A}={\mathrm{\pi R}}^2$

Part (b) Step 1: Given information

A sphere's volume is equal to ${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{a}}^2$

Part (b) Step 2: Calculation of the volume of a sphere

Take an example of sphere of radius 

Consider a sphere with a radius and a centre at the origin.

The sphere equation is ${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{a}}^2$ 

Around the x-y plane, the sphere is symmetric. As a result, the upper half of the sphere multiplied by 2 can be used to calculate its volume.

$$\begin{gathered} \mathrm{V}=2{\int }_{\mathrm{x}-\mathrm{a}}^{\mathrm{x}-\mathrm{a}} {\int }_{\mathrm{y}=0}^{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}} \mathrm{zdxdy} \\ \mathrm{V}=2{\int }_{\mathrm{x}=-\mathrm{a}}^{\mathrm{x}-\mathrm{a}} {\int }_{\mathrm{y}=0}^{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}} \sqrt{{\mathrm{a}}^2-\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}\mathrm{dxdy} \left[{\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{a}}^2\right] \end{gathered}$$

Regarding the polar shape 

Put,$\mathrm{x}=\mathrm{rcos}\mathrm{\theta },\mathrm{y}=\mathrm{rsin}\mathrm{\theta }\text{ and }\mathrm{dxdy}=\mathrm{rdrd\theta }$

$0\le \mathrm{r}\le \mathrm{a}\text{ and }-\mathrm{\pi }\le \mathrm{\theta }\le \mathrm{\pi }$

$\begin{array}{r}\mathrm{V}=2{\int }_{\mathrm{\theta }=-\mathrm{\pi }}^{\mathrm{\theta }=\mathrm{\pi }} {\int }_{\mathrm{r}=0}^{\mathrm{r}=\mathrm{\theta }} \sqrt{{\mathrm{a}}^2-\left({\mathrm{r}}^2{\cos }^2\mathrm{\theta }+{\mathrm{r}}^2{\sin }^2\mathrm{\theta }\right)}\mathrm{rdrd\theta }\\ \mathrm{V}=2{\int }_{\mathrm{\theta }=-\mathrm{\pi }}^{\mathrm{\theta }=\mathrm{\pi }} {\int }_{\mathrm{r}=0}^{\mathrm{r}=\mathrm{\theta }} \sqrt{{\mathrm{a}}^2-{\mathrm{r}}^2}\mathrm{rdrd\theta }\left[{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\right]\end{array}$

Integrate in relation to r,

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

Integrate in relation to $\mathrm{\theta },$

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

As a result, a sphere's volume is $\mathrm{V}=\frac{4}{3}{\mathrm{\pi a}}^3$

Part (c) Step 1: Given information

The double integral value is $\mathrm{A}={\mathrm{\pi r}}^2$

Part (c) Step 2: Calculation of the volume of a sphere

Let's have a look at an example of double integration.

$\mathrm{I}={\int }_0^2 {\int }_0^{\sqrt{2\mathrm{x}-{\mathrm{x}}^2}} \left({\mathrm{x}}^2+{\mathrm{y}}^2\right)\mathrm{dydx}$

The top limit of y in this case is

$\mathrm{y}=\sqrt{2\mathrm{x}-{\mathrm{x}}^2}$

${\mathrm{y}}^2=2\mathrm{x}-{\mathrm{x}}^2$ [Both squaring on sides]

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}=0...\left(\mathrm{i}\right)$

A circle with a centre is represented by Equation (1) (1,0) and the radius of a unit

 
y has a lower limit of 0

The upper part of the circle is the integration region.

Putting $\mathrm{x}=\mathrm{rcos}\mathrm{\theta },\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$  in equating (i)

$\begin{array}{r}{\mathrm{r}}^2-2\mathrm{rcos}\mathrm{\theta }=0\Rightarrow \mathrm{r}(\mathrm{r}-2\cos \mathrm{\theta })=0\\ \mathrm{r}=0,\mathrm{r}=2\cos \mathrm{\theta }\\ \mathrm{\theta }=0,\mathrm{\theta }=\frac{\mathrm{\pi }}{2}\end{array}$
That is,

$\begin{array}{r}\mathrm{I}={\int }_0^{\mathrm{\pi }/2} {\int }_0^{2\cos \mathrm{\theta }} \left({\mathrm{r}}^2{\cos }^2\mathrm{\theta }+{\mathrm{r}}^2{\sin }^2\mathrm{\theta }\right)\mathrm{rdrd\theta }\\ \mathrm{I}={\int }_0^{\mathrm{\pi }/2} {\int }_0^{2\cos \mathrm{\theta }} {\mathrm{r}}^3\mathrm{drd\theta }\\ \mathrm{I}={\int }_0^{\mathrm{\pi }/2} {\left[\frac{{\mathrm{r}}^4}{4}\right]}_0^{2\cos \mathrm{\theta }}\mathrm{d\theta }\\ \mathrm{I}={\int }_0^{\mathrm{\pi }/2} \left[\frac{16}{4}{\cos }^4\mathrm{\theta }-0\right]\mathrm{d\theta }\\ \mathrm{I}=4{\int }_0^{\mathrm{\pi }/2} {\cos }^4\mathrm{\theta d\theta }\\ \mathrm{I}=4{\int }_0^{\mathrm{\pi }/2} {\left({\cos }^2\mathrm{\theta }\right)}^2\mathrm{d\theta }\\ \mathrm{I}=4{\int }_0^{\mathrm{\pi }/2} {\left(\frac{1+\cos 2\mathrm{\theta }}{2}\right)}^2\mathrm{d\theta }\end{array}$