How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 16 - x^{2}$ $y = 16 - x^2$ and $0 \leq x \leq 4$ $0 ≤ x ≤ 4$ rotated about the x-axis?

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 16 - x^{2}$ $y = 16 - x^2$ and $0 \leq x \leq 4$ $0 ≤ x ≤ 4$ rotated about the x-axis?

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Answer 1 · 2016-10-28T09:24:39

The volume is 546.1

Explanation:

The volume of a shell is surface area multiplied by thickness
The surface area is $π y^{2}$ $piy^2$
And the volume is =surface area x thickness

$d V = π y^{2} d x$ $dV=piy^2dx$
$y = 16 - x^{2}$ $y=16-x^2$
so $y^{2} = 256 - 32 x^{2} + x^{4}$ $y^2=256-32x^2+x^4$

The volume is $V = π \int_{0}^{4} (256 - 32 x^{2} + x^{4}) d x$ $V=piint_0^4(256-32x^2+x^4)dx$

$V = π {(256 x - \frac{32 x^{3}}{3} + \frac{x^{5}}{5})}_{0}^{4}$ $V=pi(256x-(32x^3)/3+x^5/5)_0^4$
$V = π (256 \cdot 4 - 32 \cdot \frac{64}{3} + \frac{1024}{5})$ $V=pi(256*4-32*64/3+1024/5)$
$V = π (1024 - \frac{2048}{3} + \frac{1024}{5}) = 546.1$ $V=pi(1024-2048/3+1024/5)=546.1$

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 16 - x^{2}$ $y = 16 - x^2$ and $0 \leq x \leq 4$ $0 ≤ x ≤ 4$ rotated about the x-axis?

1 Answer

Explanation:

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y=16−x2y = 16 - x^2 and 0≤x≤40 ≤ x ≤ 4 rotated about the x-axis?

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Explanation:

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How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = 16 - x^{2}$ $y = 16 - x^2$ and $0 \leq x \leq 4$ $0 ≤ x ≤ 4$ rotated about the x-axis?