A region is bounded by the $y = 0$, $x = 1$ and $y = arctan(x)$. This region is rotated around a vertical line $x = b$, where $b > 1$. The solid formed has $Volume = 5$. To solve the problem, I must find the value of $b$.

I tried two different methods of solving this problem.

First, I tried using the washer method:

$$5 = π\int_0^A [(b-tan(y))^2 - (b-1)^2] dy$$

Where $A$ is the intersection between $y = arctan(x)$ and $x = 1$

I also tried the cylindrical shell method:

$$5 = 2π\int_0^1 (b-x)arctan(x)dx$$

I can't figure out how to solve for $b$ in either of these methods. How can I solve this problem?

  • $\begingroup$ For the second one you can distribute to get two integrals, one of arctanx and another of x*arctanx. The first is relatively well known and the second should fall to integration by parts.. Remember b is a constant $\endgroup$ Jan 10 '16 at 18:12

Hint Use the second form, distribute, and then use integration by parts on the second integral. Remember that $$\int \arctan x = x\arctan x - \frac 12 \log(x^2 + 1)$$ And that $b$ is a constant that can be brought outside the integral.


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