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:

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.