# Thread: Rotation aorund x-axis (volume)

1. ## Rotation aorund x-axis (volume)

Ok, this one is stumping me, I keep getting negative volume.

find the volume of the solid generated by revolving the region enclosed by the triangle with verticies (2,2) (2,4) (8,4) about the x-axis.

So I made a picture, found r(y)=2 and R(y)=y/3 (I did rise over run for the line, and solved in terms of y)
so i know its the integral $(pi) (R(y)^2)-(r(y)^2)$

found my bounds, 2 to 4, did the algebra, found anti deriv. and evaluated on the bounds, but I got $-160pi/27$ $units^3$

2. Originally Posted by BCHurricane89
Ok, this one is stumping me, I keep getting negative volume.

find the volume of the solid generated by revolving the region enclosed by the triangle with verticies (2,2) (2,4) (8,4) about the x-axis.

So I made a picture, found r(y)=2 and R(y)=y/3 (I did rise over run for the line, and solved in terms of y)
so i know its the integral $(pi) (R(y)^2)-(r(y)^2)$

found my bounds, 2 to 4, did the algebra, found anti deriv. and evaluated on the bounds, but I got $-160pi/27$ $units^3$
1. The line y = 4 (from x = 2 to x = 8) rotating around the x-axis generates a cylinder.

2. The line $y = \dfrac13x + \dfrac43$ (from x = 2 to x = 8) generates a frustrum which is taken off from the cylinder.

3. Therefore the volume is:

$V_{rot}=\int_2^8\left(\pi (4)^2 - \pi\left(\dfrac13x + \dfrac43 \right)^2 \right)dx$

I've got $V=40\pi$

3. Thanks, I see where you got that, but I just made a huge mistake, it should be around the y-axis...dooH!

4. Originally Posted by BCHurricane89
Thanks, I see where you got that, but I just made a huge mistake, it should be around the y-axis...dooH!
Then the volume is:

$V-{rot}=\int_2^4\left(\pi\left(3y-4\right)^2-\pi(2)^2 \right)dy$

I've got $V=48\pi$