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Chapter 13: Q. 39 (page 1027)

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in ${鈩�}^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
39. $2 {鈭�}_{0}^{2 蟺} {鈭�}_{0}^{4} r \sqrt{16 - r^{2}} d r d 胃$

Short Answer

Expert verified

The value of the integral is

${鈭�}_{0}^{2 蟺} {鈭�}_{0}^{4} r \sqrt{16 - r^{2}} d r d 胃 = \frac{256 蟺}{3}$

Step by step solution

01

Given information

Given expression:

$I = 2 {鈭�}_{0}^{2 蟺} {鈭�}_{0}^{4} r \sqrt{16 - r^{2}} d r d 胃$

02

Using polar coordinates to describe the solid, and evaluating the expressions.

Here, $r = 0, r = 4$ and $胃 = 0, 胃 = 2 蟺$

To integrate with respect to $r$ first,

Put $16 - r^{2} = t^{2}$

$- 2 r d r = 2 t d t r d r = - t d t r = 0 鈬� t = 4 and r = 4 鈬� t = 0$

So integral $I$ becomes:

$I = 2 {鈭�}_{0}^{2 蟺} [{鈭�}_{4}^{0} \sqrt{t^{2}} (- t d t)] d 胃 [{鈭�}_{0}^{b} f (x) d x = - {鈭�}_{0}^{a} f (x) d x] I = 2 {鈭�}_{0}^{2 蟺} [{鈭�}_{0}^{4} t^{2} d t] d 胃 I = 2 {鈭�}_{0}^{2 蟺} {[\frac{t^{3}}{3}]}_{0}^{4} d 胃 I = 2 {鈭�}_{0}^{2 蟺} [\frac{4^{3} - 0}{3}] d 胃 I = 2 {鈭�}_{0}^{2 蟺} \frac{64}{3} d 胃 I = \frac{128}{3} {鈭�}_{0}^{2 蟺} d 胃 I = \frac{128}{3} [胃]_{0}^{2 蟺} \frac{256 蟺}{3}$

Thus, the value of integral is

$2 {鈭�}_{0}^{2 蟺} {鈭�}_{0}^{4} r \sqrt{16 - r^{2}} d r d 胃 = \frac{256 蟺}{3}$

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Most popular questions from this chapter

Evaluate the sums in Exercises $23 鈥� 28$ .

{鈭�}_{j = 1}^{m} {鈭�}_{k = 1}^{n} (j - k)

Evaluate the iterated integral :

${鈭�}_{0}^{1} 鈥� {鈭�}_{0}^{3 鈭� 3 y} 鈥� {鈭�}_{0}^{6 y} 鈥� \frac{x}{y} d z d x d y$

Evaluate each of the double integrals in Exercises 37鈥�54 as iterated integrals.

$鈭� {鈭�}_{R} x \sin x \cos y d A, w h e r e R = {(x, y) | 鈭� 3 鈮� x 鈮� 2 a n d 鈭� 2 鈮� y 鈮� 2}$

Explain how to construct a Riemann sum for a function of two variables over a rectangular region.

Use the lamina from Exercise 64, but assume that the density is proportional to the distance from the x-axis.

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