How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \frac{1}{x^{4}}$ $y = 1/x^4$ , y = 0, x = 1, x = 4 revolved about the x=-4?

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \frac{1}{x^{4}}$ $y = 1/x^4$ , y = 0, x = 1, x = 4 revolved about the x=-4?

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Answer 1 · 2017-12-16T08:04:36

$\frac{57 π}{16}$ $(57pi)/16$

Explanation:

the formula for the shell method is $\int_{a}^{b} 2 π r h d x$ $int_a^b2pirhdx$

$a$ $a$ and $b$ $b$ are the x-bounds, which are x=1 and x=4, so $a = 1$ $a=1$ and $b = 4$ $b=4$ .

$r$ $r$ is the distance from a certain x-value in the interval $[1, 4]$ $[1,4]$ and the axis of rotation, which is x=-4. $r = x - (- 4) = x + 4$ $r=x-(-4)=x+4$

$h$ $h$ is the height of the cylinder at a certain x-value in the interval $[1, 4]$ $[1,4]$ , which is $\frac{1}{x^{4}} - 0 = \frac{1}{x^{4}}$ $1/x^4-0=1/x^4$ (because $\frac{1}{x^{4}}$ $1/x^4$ is always greater than $0$ $0$ and h must be positive).

plugging it all in: volume $= \int_{1}^{4} (2 π (x + 4) (\frac{1}{x^{4}})) d x$ $=int_1^4(2pi(x+4)(1/x^4))dx$
you should get: $\frac{57 π}{16}$ $(57pi)/16$

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \frac{1}{x^{4}}$ $y = 1/x^4$ , y = 0, x = 1, x = 4 revolved about the x=-4?

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Explanation:

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 1/x^4y=1x4y = 1/x^4, y = 0, x = 1, x = 4 revolved about the x=-4?

1 Answer

Explanation:

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How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = \frac{1}{x^{4}}$ $y = 1/x^4$ , y = 0, x = 1, x = 4 revolved about the x=-4?