## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 17

#### Answer

$$A = \frac{32}{3}$$

#### Work Step by Step

$$x_{1} =2y^{2}$$ $$x_{2} = 4+y^{2}$$ To find the area bounded between the two curves, we will divide the bounded region into thin horizontal strips of lenght $x_{2} \rightarrow x_{1}$ and widht $dy$. Note that the points of intersection between the two curves are $(8,-2)$ and $(8,2)$, Therefore, $y$ varies from -2 to 2. $$A = \displaystyle\int_{-2}^{2} x_{2} - x_{1} dy$$ $$=\displaystyle\int_{-2}^{2}(4+y^{2})-2y^{2}$$ $$= \left[4y - \frac{y^{3}}{3}\right]^{2}_{-2}$$ $$= \Bigg[4\times 2-\frac{2^{3}}{3}\Bigg] - \Bigg[4\times(-2)-\frac{(-2)^{3}}{3}\Bigg]$$ $$=2\Bigg[8-\frac{8}{3}\Bigg]=\frac{32}{3}$$

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