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: 34

Answer

$$V = \frac{1}{{15}}\left( {{{17}^{5/2}} - 1025} \right)$$

Work Step by Step

$$\eqalign{ & z = yx\sqrt {{x^2} + {y^2}} ;\,\,\,\,\,\,\,0 \leqslant x \leqslant 4,\,\,\,\,\,0 \leqslant y \leqslant 1 \cr & {\text{by the equation just given}}{\text{, the volume is}} \cr & V = \iint\limits_R {f\left( {x,y} \right)}dxdy \cr & {\text{replacing the limits for the region }}R{\text{ and }}f\left( {x,y} \right) = z \cr & V = \int_0^1 {\int_0^4 {yx\sqrt {{x^2} + {y^2}} } dxdy} \cr & V = \int_0^1 {\left[ {\int_0^4 {yx\sqrt {{x^2} + {y^2}} dx} } \right]dy} \cr & {\text{solve the inner integral, treating }}y{\text{ as a constant}} \cr & = \int_0^4 {yx\sqrt {{x^2} + {y^2}} } dx \cr & = \frac{1}{2}y\int_0^4 {\sqrt {{x^2} + {y^2}} \left( {2x} \right)} dx \cr & {\text{integrate by using the power rule }} \cr & = \frac{1}{2}y\left[ {\frac{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^4 \cr & = \frac{1}{3}y\left[ {{{\left( {{x^2} + {y^2}} \right)}^{3/2}}} \right]_0^4 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{1}{3}y\left[ {{{\left( {{4^2} + {y^2}} \right)}^{3/2}} - {{\left( {{0^2} + {y^2}} \right)}^{3/2}}} \right] \cr & {\text{simplifying}} \cr & = \frac{1}{3}y{\left( {16 + {y^2}} \right)^{3/2}} - \frac{1}{3}{y^4} \cr & \cr & V = \int_0^1 {\left[ {\int_0^4 {yx\sqrt {{x^2} + {y^2}} dx} } \right]dy} \cr & V = \int_0^1 {\left( {\frac{1}{3}y{{\left( {16 + {y^2}} \right)}^{3/2}} - \frac{1}{3}{y^4}} \right)dy} \cr & V = \frac{1}{3}\int_0^1 {\left( {y{{\left( {16 + {y^2}} \right)}^{3/2}} - {y^4}} \right)dy} \cr & {\text{integrating}} \cr & V = \frac{1}{3}\left[ {\frac{{{{\left( {16 + {y^2}} \right)}^{5/2}}}}{{2\left( {5/2} \right)}} - \frac{{{y^5}}}{5}} \right]_0^1 \cr & V = \frac{1}{3}\left[ {\frac{{{{\left( {16 + {y^2}} \right)}^{5/2}}}}{5} - \frac{{{y^5}}}{5}} \right]_0^1 \cr & V = \frac{1}{{15}}\left[ {{{\left( {16 + {y^2}} \right)}^{5/2}} - {y^5}} \right]_0^1 \cr & {\text{evaluating}} \cr & = \frac{1}{{15}}\left[ {{{\left( {16 + {1^2}} \right)}^{5/2}} - {1^5}} \right] - \frac{1}{{15}}\left[ {{{\left( {16 + {0^2}} \right)}^{5/2}} - {0^5}} \right] \cr & = \frac{1}{{15}}\left( {{{17}^{5/2}} - 1} \right) - \frac{1}{{15}}\left( {{{16}^{5/2}} - 0} \right) \cr & = \frac{1}{{15}}\left( {{{17}^{5/2}} - 1} \right) - \frac{1}{{15}}\left( {1024} \right) \cr & V = \frac{1}{{15}}\left( {{{17}^{5/2}} - 1 - 1024} \right) \cr & V = \frac{1}{{15}}\left( {{{17}^{5/2}} - 1025} \right) \cr} $$
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