How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

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How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

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Answer 1 · 2015-09-29T05:08:19

See the explanation below.

Explanation:

Here is a picture of the region and a vertical slice.

enter image source here

The picture is set up to use washers (disks).
Thickness is $d x$ $dx$
$x$ $x$ values go from $- 1$ $-1$ to $1$ $1$
the radius of the larger washer is the greater $y$ $y$ minus $- 2$ $-2$ (the line we are revolving about is $y = - 2$ $y=-2$ )
$R = 1 - (- 2) = 3$ $R = 1-(-2) = 3$
the radius of the smaller washer is the lesser $y$ $y$ minus $- 2$ $-2$
$r = x^{2} - (- 2) = x^{2} + 2$ $r=x^2-(-2) = x^2+2$

The representative slice has volume $π (R^{2} - r^{2}) d x$ $pi(R^2-r^2)dx$ .

We need to evaluate the integral

$\int_{- 1}^{1} π ((3^{2} - {(x^{2} + 2)}^{2}) d x = π \int_{- 1}^{1} (5 - 4 x^{2} - x^{4}) d x$ $int_-1^1 pi((3^2-(x^2+2)^2)dx=pi int_-1^1(5-4x^2-x^4)dx$

$= \frac{104}{15} π$ $= 104/15pi$

Shells Method
If we had taken a slice horizontally:

enter image source here

This is set up to use cylindrical shells of thickness $d y$ $dy$

The volume of each shell is $2 π (radius) (height) (thickness)$ $2pi("radius")("height")("thickness")$ with thickness $d y$ $dy$ , that is $2 π \cdot r h \cdot d y$ $2pi*rh*dy$
In the region, the $y$ $y$ values go from $0$ $0$ to $1$ $1$ .
The radius of the representative shell is $y + 2$ $y+2$ (dotted line in picture)
The height goes from the greater $x$ $x$ (the one on the right) to the lesser $x$ $x$ (the one on the left).
We need to rewrite the boundary as functions of $y$ $y$ instead of $x$ $x$ . $y = x^{2}$ $y=x^2$ becomes the two functions $x = - \sqrt{y}$ $x=-sqrty$ (on the left) and $x = \sqrt{y}$ $x=sqrty$ (on the right). The height of the shell is $\sqrt{y} - (- \sqrt{y}) = 2 \sqrt{y}$ $sqrty-(-sqrty) = 2sqrty$

The representative shell has volume $2 π (y + 2) (2 \sqrt{y}) d y$ $2pi(y+2)(2sqrty)dy$ .

The solid has volume

$V = \int_{0}^{1} 2 π (y + 2) (2 \sqrt{y}) d y = 4 π \int_{0}^{1} (y^{\frac{3}{2}} + 2 y^{\frac{1}{2}}) d y$ $V = int_0^1 2pi(y+2)(2sqrty)dy = 4piint_0^1 (y^(3/2)+2y^(1/2))dy$

$= 4 π (\frac{26}{15}) = \frac{104}{15} π$ $=4pi(26/15) = 104/15 pi$

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How do you find the volume of the solid generated by revolving the region bounded by the curves y = x² and y =1 rotated about the y=-2?

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