Volume of a solid formed as vertical limit goes to infinity

Question

Here's the question:

The way I have it set up currently is as follows:

$V = \pi \lim_{a \to \infty} \int_1^a (a-1)^2 - (\frac{1}{\sqrt{x^5}} - 1)^2$

But how do I go from here? And is the working thus far okay? Note that I do not have access to a graphing calculator or WolframAlpha in this scenario.

Update:

So I've determined that the $\lim_{a \to \infty} \frac{1}{\sqrt{a^5}} = 0$

and as such decided to compute this as the volume:

$V = \pi \int_0^1 (\frac{1}{\sqrt{x^5}} - 1)^2$

I ended with: $\frac{15\pi}{4}$.

Is this right?

Answer 1

You are missing the $\text{d}x$ on your integrals. Aside from that, the area is

\begin{align} \int_1^a \pi (1 - x^{-5/2})^2 \ \text{d} x &= \pi \left[ \frac{4}{3x^{3/2}} - \frac{1}{4x^4} + x \right]_1^a \\ &= \pi \left[\frac{4}{3a^{3/2}} - \frac{1}{4a^4} + a - \frac{25}{12} \right] \\ \end{align}

Hence,

$$ \lim_{a \to \infty} \int_1^a \pi (1 - x^{-2/5})^2 \ \text{d} x = \infty.$$

So, the volume diverges to infinity as $a$ tends to infinity.