## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 51

#### Answer

$$\frac{7}{3}\left( {e - 1} \right)$$

#### Work Step by Step

\eqalign{ & \iint\limits_R {{e^{x/{y^2}}}}dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \leqslant y \leqslant 2,\,\,\,\,\,\,0 \leqslant x \leqslant {y^2} \cr & {\text{Replacing the limits for the region }}R \cr & \int_1^2 {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} dy} \cr & = \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} } \right]dy} \cr & {\text{solve the inner integral treat }}y{\text{ as a constant}} \cr & = {y^2}\int_0^{{y^2}} {{e^{x/{y^2}}}\left( {\frac{1}{{{y^2}}}} \right)dx} \cr & {\text{integrate by using }}\int {{e^u}} du = {e^u} + C \cr & = {y^2}\left( {{e^{x/{y^2}}}} \right)_0^{{y^2}} \cr & {\text{evaluating the limits in the variable }}x \cr & = {y^2}\left( {{e^{{y^2}/{y^2}}} - {e^{0/{y^2}}}} \right) \cr & = {y^2}\left( {e - 1} \right) \cr & {\text{then}} \cr & \cr & \int_1^2 {\left[ {\int_0^{{y^2}} {{e^{x/{y^2}}}dx} } \right]dy} = \int_1^2 {{y^2}\left( {e - 1} \right)dy} \cr & = \left( {e - 1} \right)\int_1^2 {{y^2}dy} \cr & {\text{integrating}} \cr & = \left( {e - 1} \right)\left[ {\frac{{{y^3}}}{3}} \right]_1^2 \cr & {\text{evaluating}} \cr & = \left( {e - 1} \right)\left[ {\frac{{{2^3}}}{3} - \frac{{{1^3}}}{3}} \right] \cr & {\text{simplifying}} \cr & = \left( {e - 1} \right)\left( {\frac{8}{3} - \frac{1}{3}} \right) \cr & = \frac{7}{3}\left( {e - 1} \right) \cr}

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