# How do you find the volume of solid formed by rotating the region bounded by the graphs: y=sqrt(x)+5; y=5; x=1; and x=0; around y=2?

Dec 5, 2016

#### Explanation:

The region described is shown below in blue. A representative slice is taken by the black line inside the region.The axis of rotation is the line $y = 2$, shown as a red dashed line. The shorter radius is the black dashed line from $y = 2$ to $y = 5$ and the longer radius is the dotted green line from $y = 2$ to the curve $y = \sqrt{x} + 5$ When the slice is rotated the result will be a thin disc (washer).

Let us denote the greater radius by $R$ and the lesser by $r$.

The volume of the disc (washer) will be $\left(\pi {R}^{2} - \pi {r}^{2}\right) \cdot \text{thickness}$.

In this picture the thickness is $\mathrm{dx}$

R = (sqrtx+5)-2) = sqrtx+3

$r = 5 - 2 = 3$

So the representative disc has volume

$\pi \left({\left(\sqrt{x} + 3\right)}^{2} - {3}^{2}\right) \mathrm{dx} = \pi \left(x + 6 \sqrt{x}\right) \mathrm{dx}$

As the final step in setting up the integral, we note that $x$ (the variable in the differential $\mathrm{dx}$) takes values from $0$ to $1$

So the volume of the solid is

$V = {\int}_{0}^{1} \pi \left(x + 6 \sqrt{x}\right) \mathrm{dx} = \frac{9 \pi}{2}$