# Find the volume of the region bounded by the graphs of y=lnx, y=0, and x=4 rotated about x=5.

I graphed the region and rotated it about the line x=5 but I'm not sure which method to solving the volume I should use.

I have tried the method by cylindrical shells but end up getting a very complicated integral that I haven't learned to solve yet.

I also tried the washer method but I'm not sure what to use as the outer and inner radius. Any hints would be appreciated. Thank you.

By shells

$$2\pi\int_1^4 (5-x)\ln x \ dx$$

That will require integration by parts.

Using washers...

$$x = e^y$$

$$\pi \int_0^{\ln 4} (5-e^y)^2 - 1\ dy$$

Washers looks like the easier way to go.

But if you have done shells and want to check your work...

$$u = \ln x, dv =5-x\ dx\\ du = \frac 1x\ dx, v = 5x - \frac {x^2}2$$

$$2\pi\left((5x - \frac {x^2}2)\ln x|_1^4 - \int_1^4 (5-\frac {x}{2} \ dx\right)$$

And the rest isn't too bad from there.

Might just do it both ways to check the result.

It's always helpful to visualize problems in calculus:

Now, lets say we want the total volume of the region bounded by $$log(x)$$, $$x = 4$$ and $$y = 0$$ rotated around the axis $$x = 5$$. Well, we just note that this is essentially the volume created by rotating the region bounded by $$log(x)$$, $$x = 5$$ and $$y = 0$$ around the axis $$x = 5$$, and subtracting from this the volume created by rotating the region bounded by $$log(x)$$, $$x = 4$$, $$x = 5$$ and $$y = 0$$ around the axis $$x = 5$$.

Thus, the volume that we want is equivalent to:

\begin{align} &\Bigg[ \int_0^{ln(5)} (5 - e^y)^2 dy - \int_{ln(4)}^{ln(5)} (5 - e^y)^2 dy - 1^2ln(4) \Bigg]\pi \\ &=\Bigg[ \int_{0}^{ln(4)} (5 - e^y)^2 dy - ln(4) \Bigg]\pi \\ &= \pi \Bigg[ 25y + \frac{1}{2}e^{2y} - 10e^y\Bigg]_0^{ln(4)} - \pi ln(4) \\ &= 48\pi ln(2) - \frac{45}{2}\pi \end{align}