# Volumes of a solid of revolutions

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• Mar 10th 2010, 11:55 PM
Awsom Guy
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.
• Mar 10th 2010, 11:56 PM
VonNemo19
Quote:

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?
• Mar 10th 2010, 11:57 PM
Awsom Guy
oops sorry, 8 and 0.
• Mar 11th 2010, 12:00 AM
VonNemo19
Quote:

Originally Posted by Awsom Guy
oops sorry, 8 and 0.

So, by discs we have $V=\pi\int_0^8[x^{1/3}]^2dx$
• Mar 11th 2010, 12:01 AM
Awsom Guy
umm.. [x^(1/3)]^2, because of the volume thing. :)
• Mar 11th 2010, 12:02 AM
VonNemo19
Note the edit
• Mar 11th 2010, 12:02 AM
Awsom Guy
yes I did that so... continue please
• Mar 11th 2010, 12:04 AM
VonNemo19
Quote:

Originally Posted by Awsom Guy
yes I did that so... continue please

Evaluate $\frac{3}{5}x^{5/3}\Big|_0^8$
• Mar 11th 2010, 12:05 AM
Awsom Guy
I see my mistake thanks...
• Mar 11th 2010, 12:06 AM
Awsom Guy
how did you get 5. on the denominator.
• Mar 11th 2010, 12:08 AM
Awsom Guy
how is it power of 5/3 I think it should be 10/3.
• Mar 11th 2010, 12:10 AM
VonNemo19
Quote:

Originally Posted by Awsom Guy
how did you get 5. on the denominator.

By the power rule for integration:

$\int{x^{2/3}}dx=\frac{x^{5/3}}{5/3}+C=\frac{3}{5}x^{5/3}+C$
• Mar 11th 2010, 12:11 AM
VonNemo19
Quote:

Originally Posted by awsom guy
how is it power of 5/3 i think it should be 10/3.

2/3+1=5/3
• Mar 11th 2010, 12:11 AM
Awsom Guy
are you sure it isn't:
x^(1/3)^2
= x^(10/3).
• Mar 11th 2010, 12:13 AM
VonNemo19
Quote:

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

Yes, I'm sure.

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