Integral of volume over an interval part 2

[Kcashew]

This problem is a continuation of the one I was doing from my last thread, I need help on part B of this question.

In order to find the volume with respect to the y axis, I tried to put the equation in terms of y.

Am I on the right track by doing this?

If not, is there some formula that I am not using?

 

[MarkFL]

For this, I would use the shell method, where:

[MATH]dV=2\pi x\left(\frac{3}{\sqrt{x^2+64}}\right)\,dx[/MATH]
And so:

[MATH]V=3\pi\int_0^{15}\frac{2x}{\sqrt{x^2+64}}\,dx[/MATH]
I would next try the substitution:

[MATH]u=x^2+64\implies du=2x\,dx[/MATH]
Can you rewrite the definite integral completely in terms of \(u\)?

 

[Kcashew]

I believe the integral would be u^(-1/2)du from 289 to 64.

The anti derivative would become 2(u^1/2), and expands out to be 34 minus 16, which equals 18.

 

[Kcashew]

Is the shell method the only way to find volume around the y-axis?

I didn't mention it before, but I am unsure as to why I do not need to express this equation in terms of y.

 

[MarkFL]

I believe the integral would be u^(-1/2)du from 289 to 64.

The anti derivative would become 2(u^1/2), and expands out to be 34 minus 16, which equals 18.

I get:

[MATH]V=3\pi\int_{64}^{289} u^{-\frac{1}{2}}\,du=6\pi(17-8)=54\pi[/MATH]

Is the shell method the only way to find volume around the y-axis?

I didn't mention it before, but I am unsure as to why I do not need to express this equation in terms of y.

We could use the disk method, but it would be more difficult. Using the shell method, the thickness of the shells is \(dx\) and so we want our integrand to be in terms of \(x\). If you want to try using the disk method, I will be glad to help you do that.

 

[MarkFL]

Let's go ahead and use the disk method to check our results. Consider the following diagram:



Revolving the green area about the \(y\)-axis will just give us a cylinder, whose radius is 15 and height is [MATH]\frac{3}{17}[/MATH], and so:

[MATH]V_1=\pi(15)^2\left(\frac{3}{17}\right)=\frac{675\pi}{17}[/MATH]
To revolved the red area, we need so solve the function for \(x\):

[MATH]y=\frac{3}{\sqrt{x^2+64}}[/MATH]
[MATH]x^2=\frac{9}{y^2}-64[/MATH]
And so, using the disk method, we find:

[MATH]dV_2=\pi\left(\frac{9}{y^2}-64\right)\,dy[/MATH]
Hence:

[MATH]V_2=\pi\int_{\frac{3}{17}}^{\frac{3}{8}} 9y^{-2}-64\,dy=-\pi\left[64y+\frac{9}{y}\right]_{\frac{3}{17}}^{\frac{3}{8}}=\frac{243\pi}{17}[/MATH]
Adding the partial volumes to get the total, we obtain:

[MATH]V=V_1+V_2=\frac{675\pi}{17}+\frac{243\pi}{17}=54\pi\quad\checkmark[/MATH]