Calculus III Advanced (Course) (14.03) (Homework)

Section 14.3 Homework

From Calculus 10e by Larson and Edwards, p. 991. Exercises 6, 18, 22, 26, 30, 40, 44.

Exercise 14.3.6 Describing a Region

Use polar coordinates to describe the region shown in Figure 1.

Figure 1

Solution The region \(R\) contains all points in a circle with radius 2 and centered at \((0,2)\). In cartesian coordinates,

\(x^{2}+(y-2)^{2} \)	\(\leqslant 4 \)
\( x^{2}+(y-2)(y-2) \)	\(\leqslant 4 \)
\( x^{2}+ y^{2}-2y-2y+4 \)	\(\leqslant 4 \)
\( x^{2}+ y^{2}-4y+4 \)	\(\leqslant 4 \)
\( x^{2}+ y^{2}-4y \)	\(\leqslant 0. \)

Substituting \(x=r \cos \theta\) and \(y= r \sin \theta\) produces

\( r^{2} \cos^2 \theta + r^{2} \sin^{2} \theta -4(r \sin \theta)\)	\( \leqslant 0 \)
\( r^{2}( \cos^2 \theta + \sin^{2} \theta) -4(r \sin \theta)\)	\( \leqslant 0 \)
\( r^{2} -4r \sin \theta \)	\( \leqslant 0 \)
\( r^{2} \)	\( \leqslant 4r \sin \theta \)
\( r \)	\( \leqslant 4 \sin \theta \)

\(R=\{(r,\theta): 0 \leqslant r \leqslant 4 \sin \theta, \: 0 \leqslant \theta \leqslant 2 \pi\} \)

Exercise 14.3.18 Converting to Polar Coordinates

Evaluate the integral by converting to polar coordinates.

$$\int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} x \: dy \: dx$$

Solution The cartesian bounds are

\( 0 \leqslant x \leqslant a\)

and

\( 0 \leqslant y \leqslant \sqrt{a^{2}-x^{2}}.\)

Since

\(y \)	\(=\sqrt{a^{2}-x^{2}} \)
\(y^{2} \)	\(= a^{2}-x^{2} \)
\(x^{2}+ y^{2} \)	\(= a^{2}\)

The graph for \(f\) is a circle with radius \(a\) starting at the origin. This makes the polar bounds,

\( 0 \leqslant r \leqslant a\)

and

\(0 \leqslant \theta \leqslant \pi/2\).

$$\int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} x \: dy \: dx $$	$$=\int_{0}^{\pi/2} \int_{0}^{a } (r \: \cos \theta)r \: dr \: d\theta $$
	$$=\int_{0}^{a} \int_{0}^{2 \pi} r^{2} \: \cos \theta \: dr \: d\theta $$
	$$= \left. \int_{0}^{a} \frac{1}{3} r^{3} \cos \theta \right]_{0}^{2 \pi} \:d\theta$$
	$$= \int_{0}^{a} \frac{8}{3} \pi^{3} \cos \theta \:d\theta$$
	$$= \left. \frac{8}{3} \pi^{3} \sin \theta \right]_{0}^{a} $$
	$$= \frac{8}{3} \pi^{3} \sin (a)$$

Exercise 14.3.22 Converting to Polar Coordinates

Evaluate the iterated integral by converting to polar coordinates.

$$ \int_{0}^{2} \int_{y}^{\sqrt{8-y^{2}}} \sqrt{x^{2}+y^{2}} \: dx \:dy$$

Solution Cartesian bounds are

\( 0 \leqslant y \leqslant 2 \)

and

\( y \leqslant x \leqslant \sqrt{8-y^{2}}. \)

Now to find \(x\)'s range.

\( y \leqslant x \leqslant \sqrt{8-y^{2}} \)
\( 0 \leqslant x \leqslant \sqrt{8-0^{2}} \)
\( 0 \leqslant x \leqslant \sqrt{8} \)	Range for \(x\)

\(r^{2} \)	\(= \sqrt{x^{2}+y^{2}} \)
\(r\)	\(= x^{2}+y^{2} \)

The region \(R\) is a \(1/8\) circle centered at the origin and ranging from \(0 \leqslant \theta \leqslant \pi/4 \).

$$\int_{0}^{2} \int_{0}^{\sqrt{8}} \sqrt{x^{2}+y^{2}} \: dx \:dy $$	$$= \int_{0}^{\pi/4} \int_{0}^{2\sqrt{2}} r \: rd \: d\theta$$
	$$= \left. \int_{0}^{\pi/4} \frac{1}{2}r^{2} \right]_{0}^{2\sqrt{2}} \: d\theta$$
	$$= \int_{0}^{\pi/4} 4 \: d\theta $$
	$$= \left. \vphantom{\frac{1}{2}} 4\theta \right]_{0}^{\pi/4} = \pi $$

Exercise 14.3.26 Converting to Polar Coordinates

Evaluate the iterated integral by converting to polar coordinates

$$ \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \sin \sqrt{x^{2}+y^{2}} \: dy \: dx. $$

Solution Cartesian bounds are

\( 0 \leqslant x \leqslant 2 \)

and

\( 0 \leqslant y \leqslant \sqrt{4-x^{2}}. \)

Now to find \(\theta\)'s range. The graph is in the first quadrant so

\( 0 \leqslant \theta \leqslant \pi/2\)

and

\( 0 \leqslant r \leqslant 2. \)

Let \(r = \sqrt{x^{2}+y^{2}}.\)

$$ \int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \sin (\sqrt{x^{2}+y^{2}}) \: dy \: dx $$	$$= \int_{0}^{\pi/2} \int_{0}^{2} \sin( r) \: r \: dr \:d\theta $$
	$$= \int_{0}^{\pi/2} \int_{0}^{2}r \sin (r) \: dr \:d\theta $$
	$$= \int_{0}^{\pi/2} \sin(2)-2\cos (2) \:d\theta $$
	$$= \left. \vphantom{\frac{1}{2}} (\sin(2)-2\cos (2))\theta \right]_{0}^{\pi/2} $$
	$$= \frac{\pi}{2} (\sin(2)-2\cos (2)) $$

Exercise 14.3.30 Converting to Polar Coordinates

Use polar coordinates to set up and evaluate the double integral \(\int_{R} \int f(x,y) \: dA.\)

$$ f(x,y) = e^{-(x^{2}+y^{2})/2}$$

\( R: \: x^{2}+y^{2} \geqslant 25, \: x \geqslant 0\)

Solution Let \(r^{2}=x^{2}+y^{2}\).

$$\int \int_{R} e^{-(x^{2}+y^{2})/2} \:dA $$	$$= \int_{-\pi/2}^{\pi/2} \int_{0}^{5} e^{-r^{2}/2} \:r\: dr \:d\theta $$
	$$= \pi \int_{0}^{5} e^{-r^{2}/2} \: d \left( \frac{r^{2}}{2} \right)$$
	$$= -\pi e^{-r^{2}/2} ]_{0}^{5} $$
	$$= \pi(1-e^{-25/2})$$

Exercise 14.3.40 Volume

Use a double integral in polar coordinates to find the volume for a sphere with radius \(a\).
Solution The equation for a sphere's volume is

$$x^{2}+y^{2}+z^{2} \rightarrow z = \pm \sqrt{a^{2}- (x^{2}+y^{2})}. $$

Let \(r^{2}=x^{2}+y^{2}.\)

$$Volume$$	$$= \int_{0}^{2\pi}\int_{0}^{a} 2 \sqrt{a^{2}-r^{2}} r \: \:dr \: d\theta $$
	$$=-2 \pi \int_{0}^{a} (a^{2}-r^{2})^{1/2} d(a^{2}-r^{2}) $$
	$$= -2 \pi \frac{2}{3} \left[ (a^{2}-r^{2})^{3/2} \right]_{0}^{a}$$
	$$= \frac{4}{3} \pi a^{3} $$

Exercise 14.3.44 Area

Use a double integral to find the area for the region in Figure 2.

Figure 2

Solution

$$Area $$	$$=\int_{0}^{2\pi} \int_{0}^{2+ \sin \theta} r \: dr \: d\theta $$
	$$= \int_{0}^{2\pi} \frac{1}{2} (2+ \sin \theta)^{2} \: d\theta$$
	$$= \int_{0}^{2\pi} \left( 2+2 \sin \theta + \frac{\sin^{2} \theta}{2} \right) \: d\theta $$
	$$= \left[ \vphantom{\frac{1}{2}} 2\theta - 2 \cos \theta\right]_{0}^{2\pi} + \frac{1}{2} \int_{0}^{2\pi} \frac{1-\cos(2\theta)}{2} d\theta $$
	$$=4\pi + \frac{1}{4} \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{2\pi} $$
	$$= 4\pi + \frac{\pi}{2} = \frac{9\pi}{2}.$$

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Parent Article: Calculus III Advanced (Course)