## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 14 - Multiple Integrals - 14.3 Double Integrals In Polar Coordinates - Exercises Set 14.3 - Page 1024: 1

#### Answer

$$\frac{1}{6}$$

#### Work Step by Step

\eqalign{ & \int_0^{\pi /2} {\int_0^{\sin \theta } {r\cos \theta } } drd\theta \cr & = \int_0^{\pi /2} {\left[ {\int_0^{\sin \theta } {r\cos \theta } dr} \right]} d\theta \cr & {\text{solve the inner integral and treat }}\theta {\text{ as a constant}} \cr & = \int_0^{\sin \theta } {r\cos \theta } dr \cr & = \cos \theta \int_0^{\sin \theta } r dr \cr & {\text{then}} \cr & = \cos \theta \left[ {\frac{{{r^2}}}{2}} \right]_0^{\sin \theta } \cr & {\text{evaluating the limits in the variable }}r \cr & = \cos \theta \left[ {\frac{{{{\sin }^2}\theta }}{2} - \frac{{{0^2}}}{2}} \right] \cr & {\text{simplifying}} \cr & = \cos \theta \left[ {\frac{{{{\sin }^2}\theta }}{2} - \frac{{{0^2}}}{2}} \right] \cr & = \frac{1}{2}{\sin ^2}\theta \cos \theta \cr & {\text{then}} \cr & \int_0^{\pi /2} {\left[ {\int_0^{\sin \theta } {r\cos \theta } dr} \right]} d\theta = \int_0^{\pi /2} {\frac{1}{2}{{\sin }^2}\theta \cos \theta } d\theta \cr & {\text{integrating by the power rule}} \cr & = \frac{1}{2}\left( {\frac{{{{\sin }^3}\theta }}{3}} \right)_0^{\pi /2} \cr & {\text{evaluate}} \cr & = \frac{1}{6}\left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) - {{\sin }^3}\left( 0 \right)} \right) \cr & = \frac{1}{6}\left( {1 - 0} \right) \cr & = \frac{1}{6} \cr}

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