Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 26

Answer

$$ = \frac{{316}}{{15}} - \frac{{26\sqrt 5 }}{3}$$

Work Step by Step

$$\eqalign{ & \iint\limits_R {\frac{y}{{\sqrt {2x + 5{y^2}} }}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 2,\,\,\,\,\,\,\,1 \leqslant y \leqslant 3 \cr & \cr & {\text{replacing the limits for the region }}R \cr & \int_1^3 {\int_0^2 {\frac{y}{{\sqrt {2x + 5{y^2}} }}dx} dy} \cr & = \int_1^3 {\left[ {\int_0^2 {\frac{y}{{\sqrt {2x + 5{y^2}} }}} dx} \right]dy} \cr & {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr & = \frac{y}{2}\int_0^2 {{{\left( {2x + 5{y^2}} \right)}^{ - 1/2}}2} dx \cr & {\text{integrate by using the power rule}} \cr & = \frac{y}{2}\left[ {\frac{{{{\left( {2x + 5{y^2}} \right)}^{1/2}}}}{{1/2}}} \right]_0^2 \cr & = y\left[ {\sqrt {2x + 5{y^2}} } \right]_0^2 \cr & {\text{evaluating the limits in the variable }}x \cr & = y\left[ {\sqrt {2\left( 2 \right) + 5{y^2}} } \right] - y\left[ {\sqrt {2\left( 0 \right) + 5{y^2}} } \right] \cr & {\text{simplifying}} \cr & = y\sqrt {4 + 5{y^2}} - y\left[ {\sqrt 5 y} \right] \cr & = y\sqrt {4 + 5{y^2}} - \sqrt 5 {y^2} \cr & \cr & \int_1^3 {\left[ {\int_0^2 {\frac{y}{{\sqrt {2x + 5{y^2}} }}} dx} \right]dy} = \int_1^3 {\left( {y\sqrt {4 + 5{y^2}} - \sqrt 5 {y^2}} \right)dy} \cr & = \int_1^3 {y\sqrt {4 + 5{y^2}} dy} - \int_1^3 {\sqrt 5 {y^2}dy} \cr & = \frac{1}{{10}}\int_1^3 {10y\sqrt {4 + 5{y^2}} dy} - \sqrt 5 \int {{y^2}} dy \cr & {\text{integrating}} \cr & = \frac{1}{{10}}\left( {\frac{{{{\left( {4 + 5{y^2}} \right)}^{3/2}}}}{{3/2}}} \right)_1^3 - \sqrt 5 \left( {\frac{{{y^3}}}{3}} \right)_1^3 \cr & = \frac{1}{{15}}\left( {{{\left( {4 + 5{y^2}} \right)}^{3/2}}} \right)_1^3 - \frac{{\sqrt 5 }}{3}\left( {{y^3}} \right)_1^3 \cr & {\text{evaluate}} \cr & = \frac{1}{{15}}\left( {{{\left( {4 + 5{{\left( 3 \right)}^2}} \right)}^{3/2}} - {{\left( {4 + 5{{\left( 1 \right)}^2}} \right)}^{3/2}}} \right) - \frac{{\sqrt 5 }}{3}\left( {{{\left( 3 \right)}^3} - {{\left( 1 \right)}^3}} \right) \cr & {\text{simplifying}} \cr & = \frac{1}{{15}}\left( {{{\left( {49} \right)}^{3/2}} - {{\left( 9 \right)}^{3/2}}} \right) - \frac{{\sqrt 5 }}{3}\left( {{{\left( 3 \right)}^3} - {{\left( 1 \right)}^3}} \right) \cr & = \frac{1}{{15}}\left( {343 - 27} \right) - \frac{{\sqrt 5 }}{3}\left( {26} \right) \cr & = \frac{{316}}{{15}} - \frac{{26\sqrt 5 }}{3} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.