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Chapter 13: Q. 40 (page 1027)

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $R^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$2 {鈭�}_{0}^{2 蟺} 鈥� {鈭�}_{0}^{R} 鈥� r \sqrt{R^{2} 鈭� r^{2}} 诲谤诲胃$

Short Answer

Expert verified

The value of integral is

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

Step by step solution

01

Given information

The expression is

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

02

Calculation

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

To integrate with respect to $r$ first,

Put

localid="1651390046153" role="math" $R^{2} - r^{2} = t^{2} - 2 r d r = 2 t d t r d r = - t d t r = 0 鈬� t = R and r = R 鈬� t = 0$

So integral $I$ become

localid="1650634581886" $I = 2 {鈭�}_{0}^{2 蟺} [{鈭�}_{R}^{0} \sqrt{t^{2}} (- t d t)] d 胃 [{鈭�}_{0}^{b} f (x) d x = - {鈭�}_{0}^{a} f (x) d x] I = 2 {鈭�}_{0}^{2 蟺} [{鈭�}_{0}^{R} t^{2} d t] d 胃 I = 2 {鈭�}_{0}^{2 蟺} {[\frac{t^{3}}{3}]}_{0}^{R} d 胃 I = 2 {鈭�}_{0}^{2 蟺} [\frac{R^{3} - 0}{3}] d 胃 I = 2 {鈭�}_{0}^{2 x} \frac{R^{3}}{3} d 胃 I = \frac{2 R^{3}}{3} {鈭�}_{0}^{2 蟺} d 胃 I = \frac{2 R^{3}}{3} [胃]_{0}^{2 蟺} I = \frac{4}{3} 蟺 R^{3}$

Thus, the value of integral is

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

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Most popular questions from this chapter

Explain how to construct a Riemann sum for a function of three variables over a rectangular solid.

In the following lamina, all angles are right angles and the density is constant:

$How many summands are in {鈭�}_{i = 2}^{15} {鈭�}_{j = 3}^{17} {鈭�}_{k = 4}^{19} \frac{i}{j + k} ?$

Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32

$鈭� {鈭�}_{R} x y^{3} d A W h e r e R = {(x, y) | - 2 鈮� x 鈮� 2, - 1 鈮� y 鈮� 1}$

Describe the three-dimensional region expressed in each iterated integral in Exercises 35鈥�44.

${鈭�}_{- 3}^{3} {鈭�}_{- \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}} {鈭�}_{- \sqrt{9 - x^{2} - y^{2}}}^{\sqrt{9 - x^{2} - y^{2}}} f (x, y, z) d z d y d x$

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