# Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve y=sqrt(64-x^2), -1<=x<=1

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve
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6) For function
surface area obtained obtained by rotating the curve is given by,
$s=2\pi {\int }_{a}^{b}f\left(x\right)\sqrt{1+{\frac{dy}{dx}}^{2}}dx$
$y=f\left(x\right)=\sqrt{64-{x}^{2}}$
$⇒{f}^{\prime }\left(x\right)=\frac{-2x}{2\sqrt{64-{x}^{2}}}=\frac{-x}{\sqrt{64-{x}^{2}}}\phantom{\rule{0ex}{0ex}}⇒\left[{f}^{\prime }\left(x\right){\right]}^{2}=\frac{{x}^{2}}{\left(64-{x}^{2}\right)}\phantom{\rule{0ex}{0ex}}⇒1+\left({f}^{\prime }\left(x\right){\right)}^{2}=1+\frac{{x}^{2}}{64-{x}^{2}}=\frac{64}{64-{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)\cdot \sqrt{1+\left(\frac{dy}{dx}}{\right)}^{2}=\sqrt{64-{x}^{2}}\cdot \frac{8}{\sqrt{64-{x}^{2}}}=8$
$2\pi {\int }_{-1}^{1}8dx=32\pi$