Rotational Volume

Question

I have to find the volume of the region bounded by $ y= \sqrt{x-1} $, y=3, the y-axis and the x-axis rotated around y=5 I set up $\int_1^{10} $ $\pi((5-(\sqrt{x-1}))^2 - (5-3)^2)$dx + $\int_{0}^1$ $\pi(5^2 - (5-3)^2)$

My question is do I square each part of the integral separately like ((5^-$\sqrt{x-1})^2$) or with the whole thing squared like ((5^-$\sqrt{x-1})$)- $(5-3))^2)$?

Answer 1

You have to square them separately. What you are using is the Washer method. You are dividing the volume into thin washer like regions stacked up along the $y=5$.

The area of a washer with inner radius $r$ and outer radius $R$ is given by $$Area=\pi(R^2-r^2)$$

Here, $r$ and $R$ are the distances of $y=5$ from the functions $y=3$ and $y=0$ when $x$ ranges from $0$ to $1$ and the distances from $y=3$ and $y=\sqrt{x-1}$ when $x$ varies from $1$ to $10$ respectively. The thickness is given by $dx$, and integrating over $x$ gives the volume.

Thus,

$$Volume=\int_1^{10} \pi((5-(\sqrt{x-1}))^2 - (5-3)^2)dx + \int_{0}^1\pi(5^2 - (5-3)^2)dx$$

Answer 2

If you use the shell method instead, you will get

$\displaystyle V=\int_0^32\pi r(y)h(y)dy=\int_0^32\pi(5-y)(y^2+1)dy$.

(Here $r(y)$ is the distance from $y=5$ to a horizontal line segment in the region,

and $h(y)$ is the length of a horizontal line segment in the region.)