Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.2 Double Integrals over More General Regions - Exercises - Page 859: 29


The iterated integral after changing the order of integration: $\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y{\rm{d}}x$ Evaluate: $\mathop \smallint \limits_{x = 0}^2 \left( {\mathop \smallint \limits_{y = 0}^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y} \right){\rm{d}}x = \frac{{152}}{{15}}$

Work Step by Step

We have $\mathop \smallint \limits_0^4 \mathop \smallint \limits_{\sqrt y }^2 \sqrt {4{x^2} + 5y} {\rm{d}}x{\rm{d}}y$. From the limits of the integrals we obtain the domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 4,\sqrt y \le x \le 2} \right\}$ Notice that this is a horizontally simple region. Using this description we sketch the the domain of integration. Please see the figure attached. We change the order of integration such that the domain becomes a vertically simple region. In this case, the left and right boundaries are $x=0$ and $x=2$, respectively. Whereas, the lower boundary is $y=0$. To find the upper boundary, we use $x = \sqrt y $ and get $y = {x^2}$. So, the new domain description: ${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 2,0 \le y \le {x^2}} \right\}$ So, the iterated integral after changing the order of integration: $\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y{\rm{d}}x$ Next, we evaluate this integral: $\mathop \smallint \limits_{x = 0}^2 \left( {\mathop \smallint \limits_{y = 0}^{{x^2}} \sqrt {4{x^2} + 5y} {\rm{d}}y} \right){\rm{d}}x$ $ = \mathop \smallint \limits_{x = 0}^2 \left( {\frac{2}{{15}}{{\left( {4{x^2} + 5y} \right)}^{3/2}}|_0^{{x^2}}} \right){\rm{d}}x$ $ = \frac{2}{{15}}\mathop \smallint \limits_{x = 0}^2 \left( {{{\left( {9{x^2}} \right)}^{3/2}} - {{\left( {4{x^2}} \right)}^{3/2}}} \right){\rm{d}}x$ $ = \frac{2}{{15}}\mathop \smallint \limits_{x = 0}^2 \left( {27{x^3} - 8{x^3}} \right){\rm{d}}x$ $ = \frac{{38}}{{15}}\cdot\frac{1}{4}{x^4}|_0^2$ $ = \frac{{152}}{{15}}$
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