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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.

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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

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

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  • $\begingroup$ Thanks for the response. For the first one, how do you know the radius is y-5? $\endgroup$ – przm Mar 6 '16 at 22:59
  • $\begingroup$ 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. $\endgroup$ – user84413 Mar 6 '16 at 23:18
  • $\begingroup$ 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. $\endgroup$ – przm Mar 6 '16 at 23:39
  • $\begingroup$ That's a good picture, and y represents the distance from the center of the rectangle to the x-axis. $\endgroup$ – user84413 Mar 6 '16 at 23:55
  • $\begingroup$ Ahhh okay that makes sense - thank you! $\endgroup$ – przm Mar 7 '16 at 0:42

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