Find volume of solid generated (Calc 2)

[lovex25]

[solved]Find volume of solid generated (Calc 2)

Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve y=e^x, and the line x = ln 2 about the line x= ln 2.

So I tried graphing it to see visually, and the expression I got for calculating the volume was ∫π(ln2-lny)^2dy, evaluating from 0 to 2 using disk method, and the answer I got was 4π, but apparently that doesn't match the answer in the back of the book. I'd really appreciate if someone can help me out!

http://img844.imageshack.us/img844/9792/xll8.jpg

Off topic: First time posting a thread here, may I ask how do you type the mathematical symbols such as the integral sign and whatnot, or do I have to manually copy and paste from other website?

 

[MarkFL]

Hello lovex25,

As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:

https://www.physicsforums.com/attachments/1349._xfImport

Using the shell method, the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=\ln(2)-x\)

\(\displaystyle h=e^x\)

and so we find:

\(\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx\)

Now, you want to sum up the shells by integrating:

\(\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx\)

To make things a bit simpler, I would use the substitution:

\(\displaystyle u=e^x\,\therefore\,du=e^x\,dx\)

and we now may state:

\(\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du\)

Can you proceed?

Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:

http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html

 

[lovex25]

Hello lovex25,

As you did, a good first step for these problems is to sketch a graph of the region to be revolved, and the axis of rotation:

View attachment 1349

Using the shell method, the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=\ln(2)-x\)

\(\displaystyle h=e^x\)

and so we find:

\(\displaystyle dV=2\pi \left(\ln(2)-x \right)e^x\,dx\)

Now, you want to sum up the shells by integrating:

\(\displaystyle V=2\pi\int_0^{\ln(2)}\left(\ln(2)-x \right)e^x\,dx\)

To make things a bit simpler, I would use the substitution:

\(\displaystyle u=e^x\,\therefore\,du=e^x\,dx\)

and we now may state:

\(\displaystyle V=2\pi\int_1^2\ln(2)-\ln(u)\,du\)

Can you proceed?

Now, as far as posting mathematical expressions, this site supports $\LaTeX$, and a good tutorial written by our own Sudharaka can be found here:

http://mathhelpboards.com/latex-tips-tutorials-56/math-help-boards-latex-guide-pdf-1142.html

thanks so much! finally i was able to match the answer to the back of the book, and I realized what went wrong with my attempt: I didn't separate the integrals to 2 different functions. As for the LaTeX software I'll look into it soon, thanks again!

 

[Prove It]

I personally prefer the discs method. Here you would need to split up your region of integration into two regions, the first being the rectangle below the point (0,1), and the second being the remaining region above it.

As for the volume of the region below (0,1) generated when rotating, that's easy, it's simply a cylinder of radius [tex]\displaystyle \begin{align*} \ln{(2)} \end{align*}[/tex] units and height 1 unit, so its volume is [tex]\displaystyle \begin{align*} \pi \left[ \ln{(2)} \right] ^2 \end{align*}[/tex].

As for the volume of the region above (0,1), if we consider horizontal discs, they will each have radius [tex]\displaystyle \begin{align*} \ln{(2)} - \ln{(y)} \end{align*}[/tex] and a height [tex]\displaystyle \begin{align*} \Delta y \end{align*}[/tex], where [tex]\displaystyle \begin{align*} \Delta y \end{align*}[/tex] is some small change in y. So their total volume can be approximated by

[tex]\displaystyle \begin{align*} V &\approx \sum \pi \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left[ \ln{(2)} - \ln{(y)} \right] ^2 \Delta y \\ &= \pi \sum \left\{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \right\} \Delta y \end{align*}[/tex]

And then as we increase the number of discs, making [tex]\displaystyle \begin{align*} \Delta y \end{align*}[/tex] smaller, the sum converges on an integral and the approximation becomes exact, so the total volume is

[tex]\displaystyle \begin{align*} V &= \pi \int_1^2{ \left[ \ln{(2)} \right] ^2 - 2\ln{(2)}\ln{(y)} + \left[ \ln{(y)} \right] ^2 \, dy }\end{align*}[/tex]

which is possible to be integrated :)