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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

2.What is the volume if the above area is rotated about the line given by x=4.

Thank you in advance

- Thread starter renyikouniao
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- Thread starter
- #1

- Jun 1, 2013

- 41

2.What is the volume if the above area is rotated about the line given by x=4.

Thank you in advance

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- #2

- Jan 26, 2012

- 4,198

For both parts, I would recommend shells with a $dx$. What progress have you made?

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- #3

- Jun 1, 2013

- 41

1) my equation is:inte(lower bound: (pi/2)^0.5;upper bound: (3pi/2)^0.5)2pix*(cosx^2)

2) my equation is:inte(lower bound: (pi/2)^0.5;upper bound: (3pi/2)^0.5)2pi(4-x)*(cosx^2)

I got -2pi on the frist part,can volume be negative value?

And I got 8.37 on the second part.

2) my equation is:inte(lower bound: (pi/2)^0.5;upper bound: (3pi/2)^0.5)2pi(4-x)*(cosx^2)

I got -2pi on the frist part,can volume be negative value?

And I got 8.37 on the second part.

Last edited:

I agree with the integral you have set up for the first and the answer given by the integral. Did you notice that \(\displaystyle \displaystyle \begin{align*} \cos{ \left( x^2 \right) } < 0 \end{align*}\) when \(\displaystyle \displaystyle \begin{align*} \sqrt{ \frac{\pi}{2} } < x < \sqrt{ \frac{3\pi}{2} } \end{align*}\)? It makes sense that you would then get a negatively-signed answer for your integral. But since you are asked for the volume, you would need to take the absolute value, as you have already established that there is no such thing as a "negative volume". Also don't forget to write \(\displaystyle \displaystyle \begin{align*} \textrm{units}\,^3 \end{align*}\) after.1) my equation is:inte(lower bound: (pi/2)^0.5;upper bound: (3pi/2)^0.5)2pix*(cosx^2)

2) my equation is:inte(lower bound: (pi/2)^0.5;upper bound: (3pi/2)^0.5)2pi(4-x)*(cosx^2)

I got -2pi on the frist part,can volume be negative value?

And I got 8.37 on the second part.

- Thread starter
- #5

- Jun 1, 2013

- 41

I agree with the integral you have set up for the first and the answer given by the integral. Did you notice that \(\displaystyle \displaystyle \begin{align*} \cos{ \left( x^2 \right) } < 0 \end{align*}\) when \(\displaystyle \displaystyle \begin{align*} \sqrt{ \frac{\pi}{2} } < x < \sqrt{ \frac{3\pi}{2} } \end{align*}\)? It makes sense that you would then get a negatively-signed answer for your integral. But since you are asked for the volume, you would need to take the absolute value, as you have already established that there is no such thing as a "negative volume". Also don't forget to write \(\displaystyle \displaystyle \begin{align*} \textrm{units}\,^3 \end{align*}\) after.

integral[2*Pi*(4-x)*Cos[x^2], x, Sqrt[Pi/2], Sqrt[3*Pi/2]] - Wolfram|Alpha