# Thread: Volumes of a solid of revolutions

1. ## Volumes of a solid of revolutions

Hello,
This question I am having trouble with.
Question:
Find the volume of the solid generated when the curve y=x^(1/3) is rotaded about.

My attempt:
Intergrated = x^(10/3)
pi(8)^(10/3)
=1024pi
=3216.991
I am not sure if the answer because I think it's wrong.

2. Originally Posted by Awsom Guy
Hello,
This question I am having trouble with.
Question:
Find the volume of the solid generated when the curve y=x^(1/3) is rotaded about.

My attempt:
Intergrated = x^(10/3)
pi(8)^(10/3)
=1024pi
=3216.991
I am not sure if the answer because I think it's wrong.
What are your limits of integration?

3. oops sorry, 8 and 0.

4. Originally Posted by Awsom Guy
oops sorry, 8 and 0.
So, by discs we have $\displaystyle V=\pi\int_0^8[x^{1/3}]^2dx$

5. umm.. [x^(1/3)]^2, because of the volume thing.

6. Note the edit

7. yes I did that so... continue please

8. Originally Posted by Awsom Guy
yes I did that so... continue please
Evaluate $\displaystyle \frac{3}{5}x^{5/3}\Big|_0^8$

9. I see my mistake thanks...

10. how did you get 5. on the denominator.

11. how is it power of 5/3 I think it should be 10/3.

12. Originally Posted by Awsom Guy
how did you get 5. on the denominator.
By the power rule for integration:

$\displaystyle \int{x^{2/3}}dx=\frac{x^{5/3}}{5/3}+C=\frac{3}{5}x^{5/3}+C$

13. Originally Posted by awsom guy
how is it power of 5/3 i think it should be 10/3.
2/3+1=5/3

14. are you sure it isn't:
x^(1/3)^2
= x^(10/3).

15. Originally Posted by Awsom Guy
are you sure it isn't:
x^(1/3)^2
= x^(10/3).
Yes, I'm sure.

$\displaystyle [x^{1/3}]^2=x^{2/3}$

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