# Finding the volume of a solid bounded by curves.

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method.

$$x = (y − 9)^2, x = 16; \text{about } y = 5$$

I used the washer method in terms of y and got

$$V =\pi\int_5^{13} 16^2 - (y-9)^2 dy = \frac{8192\pi }{5} \text{ which is wrong}$$

Also, I am having problems with another similar problem:

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

$$x = 1 − y^4, x = 0, \text{ about the line } x = 5$$

Any help on how to properly set up these integrals would be great, thank you.

If you want to use the washer method for the first problem, you have to solve for y in terms of x and then integrate with respect to x, since you are revolving around a horizontal line.

It's easier to use the shell method, which gives

$\displaystyle V=\int_5^{13} 2\pi r(y)h(y) \;dy=\int_5^{13} 2\pi (y-5)(16-(y-9)^2)\;dy$

For the second problem, you could use the washer method to get

$\displaystyle V=\int_{-1}^{1}\pi\left((R(y))^2-(r(y))^2\right)dy=\int_{-1}^{1}\pi\left(5^2-(5-(1-y^4))^2\right)dy=2\int_0^1\pi\left(5^2-(4+y^4)^2\right)dy$

• Thanks for the response. For the first one, how do you know the radius is y-5? – przm Mar 6 '16 at 22:59
• If you draw a horizontal line segment in the region, corresponding to a y-value, the distance from the line segment to the line $y=5$ is the difference in the y-coordinates, which gives y-5. – user84413 Mar 6 '16 at 23:18
• I drew a horizontal rectangle and tried to visualize it, but I'm not seeing it. Is this right? i.imgur.com/UZlNxpK.png? Also, is "y" portraying the distance from 5 to y, or from the origin to y? If the latter, it makes sense why you would subtract it. – przm Mar 6 '16 at 23:39
• That's a good picture, and y represents the distance from the center of the rectangle to the x-axis. – user84413 Mar 6 '16 at 23:55
• Ahhh okay that makes sense - thank you! – przm Mar 7 '16 at 0:42