Disk Method: The bounded region formed by y=x^2, x = 3, and the x-axis is rotated about the line x = 3. Find the volume of the solid.
I believe this is a dy problem
limits are 0 to 9 correct?
yes
\[\pi \int\limits_{0}^{9}\left[ \sqrt{y}-3 \right]^2 dy\]
did I set it up correctly?
|dw:1393799102155:dw|
-9pi I believe is the solution
volumue can't be negative
what is the radius?
I dropped off the second power, but you are correct
ok I got 42.4115 I just need to integrate it by hand
your set up is correct. Technically it should be V = pi (3 - sqrt(y))^2 dy, from 0 to 9 but since the difference is squared, the answer will still come out positive
Is it right - left or left - right?
I have my functions switched above
I believe it is right - left. I guess I still don't know my right from my left (my mistake)
yeah, right - left
no wonder I washed out of the military. haha
:D
13.5 pi or 42.411 got it Thanks
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