$y=\sqrt{64-{x}^{2}},\text{}\text{}-1\le x\le 1$

Bergsteinj0
2022-09-27
Answered

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve

$y=\sqrt{64-{x}^{2}},\text{}\text{}-1\le x\le 1$

$y=\sqrt{64-{x}^{2}},\text{}\text{}-1\le x\le 1$

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Johnathon Mcmillan

Answered 2022-09-28
Author has **7** answers

6) For function $y=f(x),\text{}\text{}\text{}a\le x\le b$

surface area obtained obtained by rotating the curve is given by,

$s=2\pi {\int}_{a}^{b}f(x)\sqrt{1+{\frac{dy}{dx}}^{2}}dx$

$y=f(x)=\sqrt{64-{x}^{2}}$

$\Rightarrow {f}^{\prime}(x)=\frac{-2x}{2\sqrt{64-{x}^{2}}}=\frac{-x}{\sqrt{64-{x}^{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow [{f}^{\prime}(x){]}^{2}=\frac{{x}^{2}}{(64-{x}^{2})}\phantom{\rule{0ex}{0ex}}\Rightarrow 1+({f}^{\prime}(x){)}^{2}=1+\frac{{x}^{2}}{64-{x}^{2}}=\frac{64}{64-{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow f(x)\cdot \sqrt{1+(\frac{dy}{dx}}{)}^{2}=\sqrt{64-{x}^{2}}\cdot \frac{8}{\sqrt{64-{x}^{2}}}=8$

Answer:

$2\pi {\int}_{-1}^{1}8dx=32\pi $

surface area obtained obtained by rotating the curve is given by,

$s=2\pi {\int}_{a}^{b}f(x)\sqrt{1+{\frac{dy}{dx}}^{2}}dx$

$y=f(x)=\sqrt{64-{x}^{2}}$

$\Rightarrow {f}^{\prime}(x)=\frac{-2x}{2\sqrt{64-{x}^{2}}}=\frac{-x}{\sqrt{64-{x}^{2}}}\phantom{\rule{0ex}{0ex}}\Rightarrow [{f}^{\prime}(x){]}^{2}=\frac{{x}^{2}}{(64-{x}^{2})}\phantom{\rule{0ex}{0ex}}\Rightarrow 1+({f}^{\prime}(x){)}^{2}=1+\frac{{x}^{2}}{64-{x}^{2}}=\frac{64}{64-{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow f(x)\cdot \sqrt{1+(\frac{dy}{dx}}{)}^{2}=\sqrt{64-{x}^{2}}\cdot \frac{8}{\sqrt{64-{x}^{2}}}=8$

Answer:

$2\pi {\int}_{-1}^{1}8dx=32\pi $

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