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

[renyikouniao]

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]

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

 

[renyikouniao]

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.

 

[Prove It]

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 [math]\displaystyle \begin{align*} \cos{ \left( x^2 \right) } < 0 \end{align*}[/math] when [math]\displaystyle \begin{align*} \sqrt{ \frac{\pi}{2} } < x < \sqrt{ \frac{3\pi}{2} } \end{align*}[/math]? 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 [math]\displaystyle \begin{align*} \textrm{units}\,^3 \end{align*}[/math] after.

 

[renyikouniao]

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 [math]\displaystyle \begin{align*} \cos{ \left( x^2 \right) } < 0 \end{align*}[/math] when [math]\displaystyle \begin{align*} \sqrt{ \frac{\pi}{2} } < x < \sqrt{ \frac{3\pi}{2} } \end{align*}[/math]? 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 [math]\displaystyle \begin{align*} \textrm{units}\,^3 \end{align*}[/math] after.

 

[Prove It]

I agree with the integral you have set up. I don't agree with the answer (for one thing, you should be expecting a negative answer again which you then take the absolute value of).

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