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Question Number 98214 by me2love2math last updated on 12/Jun/20

Commented by mr W last updated on 12/Jun/20

$$A=\frac{2}{3}\times 4\times 4=\frac{32}{3}$$

Commented by mr W last updated on 12/Jun/20

$$or$$$$-1=3-y^2$$$$y^2-4=0$$$$D=0^2+4\times 4=16$$$$a=1$$$$Area=\frac{D\sqrt{D}}{6a^2}=\frac{16\times \sqrt{16}}{6\times 1^2}=\frac{32}{3}$$

Answered by Rio Michael last updated on 12/Jun/20

$$\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{limits}\mathrm{lets}\mathrm{solve}x=3-y^2\mathrm{and}x=-1$$$$\Rightarrow -1=3-y^2\iff y=\pm 2.$$$$\mathrm{the}\mathrm{required}\mathrm{area}A={\int }_{-2}^2\left(\mathrm{right}\mathrm{function}\right)-\left(\mathrm{left}\mathrm{function}\right)$$$$\Rightarrow A={\int }_{-2}^2(3-y^2-(-1)dy={\int }_{-2}^2(4-y^2)dy={[4y-\frac{y^3}{3}]}_{-2}^2$$$$=(8-\frac{8}{3})-(-8+\frac{8}{3})=16\mathrm{square}\mathrm{units}.$$$$\mathcal{R}\mathrm{ecall}\mathrm{for}\mathrm{this}\mathrm{type}\mathrm{of}\mathrm{problems}\mathrm{take}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{right}\mathrm{function}$$$$-\mathrm{the}\mathrm{left}\mathrm{function}.$$$$$$

Answered by MJS last updated on 12/Jun/20

$$x=3-y^2\iff y=\pm \sqrt{3-x}$$$$2\underset{-1}{\stackrel{3}{\int }}\sqrt{3-x}dx=-\frac{4}{3}{\left[{(3-x)}^{3/2}\right]}_{-1}^3=\frac{32}{3}$$

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