How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $y = 8 - x^{2}$ $y=8-x^2$ , $y = x^{2}$ $y=x^2$ revolved about the x=2?

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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 $y = 8 - x^{2}$ $y=8-x^2$ , $y = x^{2}$ $y=x^2$ revolved about the x=2?

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Answer 1 · 2016-11-16T21:10:20

Please see below.

Explanation:

In the following picture, the region is shown in blue. The points of intersection of the two curves are $(\pm 2, 4)$ $(+-2,4)$ . The axis of rotation (the line $x = 2$ $x=2$ ) is in red.

To set up for the method of shells, a slice has been taken parallel to the axis of rotation. The thin side (the thickness of the shell) is $d x$ $dx$ . The slice has black borders and the radius of rotation is shown as a dashed black line.

enter image source here

The Volume of a representative shell is

$2 π r h thickness$ $2 pi r h "thickness"$

Since the thickness is
$d x$ $dx$

(the variable we will work with is $x$ $x$ ) we note that $x$ $x$ varies from $- 2$ $-2$ to $2$ $2$ .

The radius of the shell is
$r = 2 - x$ $r = 2-x$ .

The height is the upper curve minus the lower curve

$h = (8 - x^{2}) - (x^{2}) = 8 - 2 x^{2}$ $h = (8-x^2) - (x^2) = 8-2x^2$

So the volume of the solid of revolution is

$V = int_-2^2 2piunderbrace((2-x))_r underbrace((8-2x^2))_h overbrace(dx)^("thickness")$

$= 2pi[128/3] = (256 pi)/3$

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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 $y = 8 - x^{2}$ $y=8-x^2$ , $y = x^{2}$ $y=x^2$ revolved about the x=2?

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=8-x^2y=8−x2y=8-x^2, y=x^2y=x2y=x^2 revolved about the x=2?

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 $y = 8 - x^{2}$ $y=8-x^2$ , $y = x^{2}$ $y=x^2$ revolved about the x=2?