Find the volume of a solid using both the disk/washer and shell method.

**tanagholat** · June 23rd 2012, 10:38 AM

Let R be the region bounded by the following curves. Let S be the solid generated  when R is revolved about the given axis. If possible, find the volume of S by both the disk/washer and shell methods. Check that your results agree.

I have attached an image with the functions and the correct answers.

I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).

I tried to type some of my work here, but when it posted it came out in a bunch of random letters and numbers and such.

Thank you for the help!

**HallsofIvy** · June 23rd 2012, 12:43 PM

I don't see why it would give difficult integrals. Rotating around the x-axis gives disks with radii y and so area $\pi y^2$ . The square root, after squaring, will be gone so you just have integrals of logarithms. What is the integral of $\int ln(ax) dx$ ?

**Reckoner** · June 24th 2012, 05:42 AM

Originally Posted by tanagholat

I have tried integrating this both as f(x) and f(y), but both lead me to have to take anti-derivatives that seem impossible with the calculus i have learned so far (u-substitution would not work).

For integrals involving logarithms, it's often useful to integrate by parts. For example, to integrate $\int\ln x^n\,dx,\ n\geq1$ let

$u = \ln x^n\Rightarrow du=\frac{nx^{n-1}}{x^n}\,dx$

$\Rightarrow du=\frac nx\,dx,$

and let $dv = dx\Rightarrow v = x,$ which gives

$\int\ln x^n\,dx = \int u\,dv$

$=uv - \int v\,du$

$=x\ln x^n - \int n\,dx$

$=x\left(\ln x^n - 1\right) + C$

Now, to the problem at hand.

Washer Method:

First we determine where $y=1$ intersects the other two curves:

$\sqrt{\ln\frac{x^2}{36}} = 1\Rightarrow x = 6\sqrt e$ (ignoring the negative root)

$\sqrt{\ln\frac x6} = 1\Rightarrow x = 6e$

So,

$V = \pi\int_6^{6\sqrt e}\left[\left(\sqrt{\ln\frac{x^2}{36}}\right)^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx + \pi\int_{6\sqrt e}^{6e}\left[1^2 - \left(\sqrt{\ln\frac x6}\right)^2\right]\,dx$

$= \pi\int_6^{6\sqrt e}\left(\ln\frac{x^2}{36} - \ln\frac x6\right)\,dx + \pi\int_{6\sqrt e}^{6e}\left(1 - \ln\frac x6}\right)\,dx$

$= \pi\left[x\left(\ln\frac{x^2}{36} - 2\right) - x\left(\ln\frac x6 - 1\right)\right]_6^{6\sqrt e} + \pi\left[x - x\left(\ln\frac x6 - 1\right)\right]_{6\sqrt e}^{6e}$

$=\pi\left[\left(6\sqrt e\right)(-1) - \left(6\sqrt e\right)\left(-\frac12\right) - 6(-2) + 6(-1) + 6e - 6e(0) - 6\sqrt e + \left(6\sqrt e\right)\left(-\frac12\right)\right]$

$=6\pi\left(e - 2\sqrt e + 1\right)$

$=6\pi\left(\sqrt e - 1\right)^2$

Shell Method:

$y = \sqrt{\ln\frac x6}\Rightarrow x = 6e^{y^2}$

$y = \sqrt{\ln\frac{x^2}{36}}\Rightarrow x = 6e^{y^2/2}$

$V = 2\pi\int_0^1y\left(6e^{y^2} - 6e^{y^2/2}\right)\,dy$

This integral is relatively straightforward. I leave it to you to show that it produces the same result.

**tanagholat** · June 24th 2012, 11:31 AM

Thank you very much to the both of you.

So we haven't learned how to do integration by parts with logarithms and such, so i guess that's where I got stuck, but other than that it's nice to know that my thought process was all correct. I did some research on how you got that and it makes some sense now. (yay)

Much appreciation to the both of you. I'm now a bit more confident with knowledge of this material.

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