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Find the volume of the solid obtained by rotating the area?

renyikouniao

Member
Jun 1, 2013
41
1.Determine the volume of the solid obtained by rotating the area between the x-axis and the graph of the function given by f(x) = cos(x^2) with x between (pi/2)^0.5 and (3pi/2)^0.5 ,about the y-axis.

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

Thank you in advance
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
For both parts, I would recommend shells with a $dx$. What progress have you made?
 

renyikouniao

Member
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.
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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.
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.
 

renyikouniao

Member
Jun 1, 2013
41
Thank you.But what about the second?Is it also correct?
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.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403