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Chapter 13: Q. 37 (page 1015)

In Exercises 35鈥�40, find the volume of the solid bounded above by the given function over the specified region $惟$ . $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$
Region:

Short Answer

Expert verified

Volume bounded by given function is $4 蟺$ .

Step by step solution

01

Step 1. Given Information

A function, $f (x, y) = \sqrt{4 - x^{2} - y^{2}}$

Region:

02

Step 2. Calculating the volume of the solid bounded above by the given function and region

The double integral is given by,

${鈭�}_{- 2}^{2} {鈭�}_{0}^{\sqrt{4 - x^{2}}} \sqrt{4 - x^{2} - y^{2}} d y d x + {鈭�}_{- 2}^{0} {鈭�}_{- \sqrt{4 - x^{2}}}^{0} \sqrt{4 - x^{2} - y^{2}} d y d x$

Due to symmetry it can also be written as,

$3 {鈭�}_{0}^{2} {鈭�}_{0}^{\sqrt{4 - x^{2}}} \sqrt{4 - x^{2} - y^{2}} d y d x$

Converting to polar coordinates, the integration becomes,

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

Put, $4 - r^{2} = t, - 2 r d r = d t,$

We get, $3 {鈭�}_{0}^{\frac{蟺}{2}} {鈭�}_{4}^{0} (- \frac{\sqrt{t}}{2}) d t d 胃$

$= - \frac{3}{2} {鈭�}_{0}^{\frac{蟺}{2}} \frac{2}{3} {(t^{\frac{3}{2}})}_{4}^{0} d 胃 = - {鈭�}_{0}^{\frac{蟺}{2}} {(0 - 4^{\frac{3}{2}})}_{4}^{0} d 胃 = - 3 {鈭�}_{0}^{\frac{蟺}{2}} \frac{2}{3} (0 - 4^{\frac{3}{2}}) d 胃 = 8 {鈭�}_{0}^{\frac{蟺}{2}} 1 d 胃 = 8 胃_{0}^{\frac{蟺}{2}} = 8 脳 \frac{蟺}{2} = 4 蟺$

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Most popular questions from this chapter

Explain why it would be difficult to evaluate the double integrals in Exercises 18 and 19 as iterated integrals.

$鈭� {鈭�}_{R} e^{x y} d A w h e r e R = \{(x, y) I 1 鈮� x 鈮� 2 a n d 1 鈮� y 鈮� 3\}$

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

${鈭�}_{0}^{2} {鈭�}_{0}^{1 - 3 x / 2} {鈭�}_{2 x + (4 / 3) y - 4}^{0} f (x, y, z) d z d y d x$

Evaluate each of the double integral in the exercise 37-54 as iterated integrals

$鈭� {鈭�}_{R} \sin (x + 2 y) d A W h e r e R = {(x, y) | 0 鈮� x 鈮� 蟺, 0 鈮� y 鈮� \frac{蟺}{2}}$

Evaluate each of the double integrals in Exercises $37 - 54$ as iterated integrals.

localid="1650380493598" $鈭� {鈭�}_{R} \sin (2 x + y) d A,$

wherelocalid="1650380496793" $R = \{(x, y) | 0 鈮� x 鈮� \frac{蟺}{2} a n d 0 鈮� y 鈮� \frac{蟺}{2}\}$

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

$鈭� {鈭�}_{R} x y d A W h e r e R = {(x, y) | 0 鈮� x 鈮� 2, 1 鈮� y 鈮� 4}$

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