How do you find the volume of the region enclosed by the curves y = x^2 - 1 and y =0 rotated around the line x = 5?

Jul 17, 2015

$V = \pi {\int}_{0}^{24} {\left(5 - \sqrt{y + 1}\right)}^{2} \mathrm{dy} = \pi \left(85 + \frac{1}{3}\right)$

Explanation:

In order to calculate this volume we are in some sense going to cut it into (infinitely slim) slices.

We envision the region, to help us with this, I have enclosed the graph where the region is the part beneath the curve. We note that $y = {x}^{2} - 1$ crosses the line $x = 5$ where $y = 24$ and that it crosses the line $y = 0$ where $x = 1$ graph{x^2-1 [1, 5, -1, 24]}

When cutting this region in horizontal slices with height $\mathrm{dy}$ (a very small height). The length of these slices depends very much on the y coordinate. to calculate this length we need to know the distance from a point $\left(y , x\right)$ on the line $y = {x}^{2} - 1$ to the point (5,y). Of course this is $5 - x$, but we want to know how it depends on $y$. Since $y = {x}^{2} - 1$, we know ${x}^{2} = y + 1$, since we have $x > 0$ for the region we are interestend in, $x = \sqrt{y + 1}$, therefore this distance dependant on $y$, which we shall denote as $r \left(y\right)$ is given by $r \left(y\right) = 5 - \sqrt{y + 1}$.

Now we rotate this region around $x = 5$, this means that every slice becomes a cylinder with height $\mathrm{dy}$ and radius $r \left(y\right)$, therefore a volume $\pi r {\left(y\right)}^{2} \mathrm{dy}$. All we need to do now is add up these infinitely small volumes using integration. We note that $y$ goes from $0$ to $24$.
$V = {\int}_{0}^{24} \pi r {\left(y\right)}^{2} \mathrm{dy} = \pi {\int}_{0}^{24} {\left(5 - \sqrt{y + 1}\right)}^{2} \mathrm{dy} = \pi {\int}_{0}^{24} \left(25 - 10 \sqrt{y - 1} + y + 1\right) \mathrm{dy} = \pi {\int}_{0}^{24} \left(26 - 10 \sqrt{y + 1} + y\right) \mathrm{dy} = \pi {\left[26 y - \frac{20}{3} {\left(y + 1\right)}^{\frac{3}{2}} + {y}^{2} / 2\right]}_{0}^{24} = \pi \left(26 \cdot 24 - \frac{20}{3} {\left(25\right)}^{\frac{3}{2}} + \frac{20}{3} + {24}^{2} / 2\right) = \pi \left(85 + \frac{1}{3}\right)$.