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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.

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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.

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It's always helpful to visualize problems in calculus:

enter image description here

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:

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

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