# Math Help - Rotate About y-axis

1. ## Rotate About y-axis

Find volume of solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

y = sqrt(x), y = 0, x = 4

ABOUT Y-AXIS.

If picture is sideways, please rotate for easy reading.

I got the right answer when rotating about the x-axis (same question). For rotation about the x-axis, my limits were [0,4] and my answer is 8pi.

Am I using the wrong limits for rotation about the y-axis or is my integral set up all wrong? If so, can someone set up the correct integral and explain how you did it. I will do the calculation on my own to find the volume.

2. ## Re: Rotate About y-axis

your limits are fine.

What you have to realise is that the shape created by the integral is not the shape created by rotating the area.

Think about it like this. at y = 0. The radius of the rotated area should be 4. But with the integral we are saying the radius is 0. vice versa.

Therefore we have to calculate the integral and subtract it from a cylinder with radius of 4 and a height of 2.

3. ## Re: Rotate About y-axis

After a second try, I got as the volume (32pi)/5. Is this correct? The textbook answer is (128pi)/5. Now, if I multiply my answer by 4, I get the textbook solution. The question is WHERE DOES 4 COME FROM?

4. ## Re: Rotate About y-axis

read above, you have to take your integral away from a cylinder the volume of the cylinder is 2*(4^2)*pi

5. ## Re: Rotate About y-axis

I sent this question to a retired math teacher who said that my answer is correct. Now I am really confused.

6. ## Re: Rotate About y-axis

did you read what I said

7. ## Re: Rotate About y-axis

I will work on this some more on my next day off.

8. ## Re: Rotate About y-axis

I need help setting up this integral.

9. ## Re: Rotate About y-axis

The outer radius is 4, and the inner radius is $y^2$, so using the washer method, the integral is:

$\displaystyle \pi \int_0^2\left(4^2-(y^2)^2\right)dy = \pi\left.\left(16y-\dfrac{y^5}{5}\right)\right]_0^2 = \dfrac{128\pi}{5}$