I need help using integrals to find volumes. If you have the equations y=9-x^2, y=0,x=0. Then you rotate the region bounded by those equations around x=3 and use integrals to get the volume of that region. I'm gettting 42.4113 for the volume but maple keeps telling me something else. Can someone tell me how to get the right answer.
Volume of y = 9 - x^2, y = 0, x = 0 rotated around x = 3
- Thread starter flash9286
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Re: Volume
Hello, flash9286!
Formula: \(\displaystyle \L\:V \;=\;2\pi\int^{\;\;\,b}_a(\text{radius})(\text{height})\,dx\)
In this problem: \(\displaystyle \,\text{radius}\:=\:3\,-\,x,\;\text{height}\:=\:9\,-\,x^2\)
So we have: \(\displaystyle \L\:V \;=\;2\pi\int^{\;\;3}_0(3\,-\,x)(9\,-\,x^2)\,dx\)
Can you finis it now?
Hello, flash9286!
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0 3Formula: \(\displaystyle \L\:V \;=\;2\pi\int^{\;\;\,b}_a(\text{radius})(\text{height})\,dx\)
In this problem: \(\displaystyle \,\text{radius}\:=\:3\,-\,x,\;\text{height}\:=\:9\,-\,x^2\)
So we have: \(\displaystyle \L\:V \;=\;2\pi\int^{\;\;3}_0(3\,-\,x)(9\,-\,x^2)\,dx\)
Can you finis it now?
You could proceed with shells or washers. Take your pick.
SHELLS:
\(\displaystyle \L\\2{\pi}\int_{0}^{9}(9-y)\sqrt{9-y}dy\)
WASHERS:
\(\displaystyle \L\\{\pi}\int_{0}^{3}(81-x^{4})dx\)
YOu should get the same result with both methods.
SHELLS:
\(\displaystyle \L\\2{\pi}\int_{0}^{9}(9-y)\sqrt{9-y}dy\)
WASHERS:
\(\displaystyle \L\\{\pi}\int_{0}^{3}(81-x^{4})dx\)
YOu should get the same result with both methods.