Find the volume of the solid obtained by rotating the region bounded by the graphs $y=\frac{1}{x}, y=0, x=1$ and $x=9$ about $ y=6 $

Question

Find the volume of the solid obtained by rotating the region bounded by the graphs $y=\frac1x,y=0,x=1$ and $x=9$ about $y=6$.

I thought that maybe by offsetting $\frac{1}{x}$ by 6 and then treating it like a normal "rotate about x-axis" that it'd work, so I set it up like this:

$$ \int_1^9 \pi\left(\frac{1}{x}-6\right)^2-6^2dx $$

but that doesn't yield the right answer for this problem according to the feedback tool. Where'd I go wrong?

Answer 1

I figured out the answer is: $$\pi\left(24 \ln (3)-\frac{8}{9}\right)$$ I wasn't thinking about the problem in the right way.

The "outer washer" is the distance from the x-axis to 6, so it's a constant. Big R, then, is 6.

The "inner washer" is the distance from $y=6$ to the function, or $6-\frac{1}{x}$.

You can then set up like: $$\pi \cdot \int_1^9 6^2-\left(6-\frac{1}{x}\right)^2 d x$$

Answer 2

You forgot to add parentheses. The correct integral is $$\int_1^9\pi(6^2-(6-1/x)^2)\,dx$$