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Chapter 13: Q. 45 (page 1079)

Evaluate the double integrals in Exercises 39鈥�48. Use suitable transformations as necessary.
$鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A$ , where $惟$ is the following region:

Short Answer

Expert verified

$鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = \frac{15}{8}$

Step by step solution

01

Draw the region and name the vertices

The region $惟$ is bounded by,

$y = 2 x, y = 2 x^{2}, y = x, y = x^{2}$

Plot the given points to form the region and name the vertices.

Consider the new set of variables defined as

$u = \frac{y}{x} v = \frac{y}{x^{2}}$

After solving ee get that,

$\frac{u}{v} = x \frac{u^{2}}{v} = y$

02

Determine the equation of each boundary in terms of u and v.

We have,

$\frac{u}{v} = x \frac{u^{2}}{v} = y$

Use these equations to determine the equation of each boundary of the region.

$A B : y = x 鈬� u = 1 B C : y = 2 x^{2} 鈬� v = 2 C D : y = 2 x 鈬� u = 2 D A : y = x^{2} 鈬� v = 1$

Plot these limits on u v plane.

03

Evaluate the double integral.

Set up the double integral.

$鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = {鈭�}_{u = 1}^{u = 2} {鈭�}_{v = 1}^{v = 2} (\frac{u^{3}}{v^{2}}) d v d u 鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = {鈭�}_{u = 1}^{u = 2} u^{3} ({鈭�}_{v = 1}^{v = 2} (\frac{1}{v^{2}}) d v) d u 鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = \frac{1}{2} {鈭�}_{u = 1}^{u = 2} u^{3} d u 鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = \frac{1}{2} {[\frac{u^{4}}{4}]}_{1}^{2} 鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = \frac{1}{2} [\frac{2^{4}}{4} - \frac{1^{4}}{4}] 鈭� {鈭�}_{惟} \frac{y^{2}}{x^{3}} d A = \frac{15}{8}$

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Most popular questions from this chapter

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

$Express the sum 3 e^{4} + 3 e^{9} + 3 e^{16} + 4 e^{4} + 4 e^{9} + 4 e^{16} using double - summation notation .$

How many summands are in ${鈭�}_{i = i_{0}}^{l} {鈭�}_{j = j_{0}}^{m} {鈭�}_{k = k_{0}}^{n} k^{i} j^{k}$ ?

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

${鈭�}_{- 3}^{3} {鈭�}_{- \sqrt{9 - z^{2}}}^{\sqrt{9 - z^{2}}} {鈭�}_{0}^{3 - \sqrt{x^{2} + z^{2}}} f (x, y, z) d y d x d z$

$L e t R = \{(x, y, z) | a_{1} 猢� x 猢� a_{2}, b_{1} 猢� y 猢� b_{2}, c_{1} 猢� z 猢� c_{2}\} . P r o v e t h a t : {鈭�}_{R} d V = (a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1}) . W h a t i s t h e r e l a t i o n s h i p b e t w e e n R a n d t h e p r o d u c t (a_{2} - a_{1}) (b_{2} - b_{1}) (c_{2} - c_{1}) .$

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