Thread: 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.
(i) about the x-axis.

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.

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.
(i) about the x-axis.

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?

oops sorry, 8 and 0.

Originally Posted by Awsom Guy

oops sorry, 8 and 0.

So, by discs we have $\displaystyle V=\pi\int_0^8[x^{1/3}]^2dx$

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

Note the edit

yes I did that so... continue please

Originally Posted by Awsom Guy

yes I did that so... continue please

Evaluate $\displaystyle \frac{3}{5}x^{5/3}\Big|_0^8$

I see my mistake thanks...

how did you get 5. on the denominator.

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

Originally Posted by Awsom Guy

how did you get 5. on the denominator.

By the power rule for integration:

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

Originally Posted by awsom guy

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

2/3+1=5/3

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

Originally Posted by Awsom Guy

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

Yes, I'm sure.

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