# volume of solid of revolution. Can someone please check my work?

Original question: Find the volume of the solid formed by revolving the region bounded by the graphs of $y=\ln x$ and $x-2y=1$ about the x-axis.

Here's what I did:

1. Get $x-2y=1$ in terms of y, therefore $y=\frac{1}{2}x-\frac{1}{2}$
2. Find intersection points for the bounds, $(1,0),(3.51286,1.25643)$
3. (Not sure how to set up integral on here, sorry) I used the washer method and came up with $\displaystyle \pi \cdot \int_1^{3.51286}\left((\ln x)^2 - (\frac{1}{2}x - \frac{1}{2})^2\right) dx$
4. ended up with $7.34579$

Did I do this correctly?

The setup is correct.

I did the numerical integral on Wolfram Alpha (see here), and got $\approx 0.421573 \pi \approx 1.3244111$ instead.

• ok thank you! I plugged it back into my calculator a second time and got 1.32 as well :) Thanks for your help – sloth1111 Jan 3 '14 at 2:12