# Finding volume by integration

Well, the question says "The area bounded by hyperbola $xy=4$ and line $x+y=5$ is revolved about $x$-axis. Find the volume of solid thus formed."

Having known that this site doesn't solve your homework problems but just helps you in understanding the concept, I will also explain what I have understood from the question and how have I tried yet got the wrong answer.

First step of mine was drawing the graphs of $xy=4$ and $x+y=5$ and found that they intersect at $(1,4)$ and $(4,1)$. Then, I expressed both expressions in terms of $x$: $y=\frac{4}{x}$ and $y=5-x$.

I end up with this integral $$\int_1^4 \pi \left((5-x)^2 - \left(\frac{4}{x}\right)^2 \right) \,dx$$

The answer I got is $93\pi$, but the book says the answer is $9\pi$. Please help me find my mistake.

$$\pi\int_1^4\left[(5-x)^2-\frac{16}{x^2}\right]dx=\pi\left(\left.-\frac13(5-x)^3\right|_1^4+\left.\frac{16}x\right|_1^4\right)=$$
$$=\pi\left(-\frac13\left[1-4^3\right]+16\left[\frac14-1\right]\right)=\pi(21-12)=9\pi$$