Volume when the region bounded by $x=y^2$ and $x=y+2$ is revolved about the $y$-axis

Question

How do you find the volume of the solid created when the region bounded by $x=y^2$ and $x=y+2$ is revolved about the $y$-axis? Thank you.

Answer 1

You need to calculate $Volume = \int_a^bdV$, where finding $dV$ is most of the work. You begin by choosing to either divide the bounded area into horizontal rectangles each with height $dy$, or vertical rectangles with width $dx$. In either case, $dV$ is the expression for the volume of the solid obtained by revolving a general rectangle (of thickness either $dx$ or $dy$) about the axis in question. This solid with volume $dV$ will be either a thin disk/washer or a thin-walled cylinder.

In this case, using horizontal rectangles will be much simpler. Rotating a horizontal rectangle about the y-axis creates a thin washer.

The bounded area lies in the domain $$-1\le y \le 2.$$ In this domain, the distance from the y-axis to the left side of one of the horizontal rectangles is $$y^2$$ and the distance from the y-axis to the right side of a rectangle is $$y+2.$$ The horizontal rectangles have a thickness (height) of $$dy.$$

So now having determined the dimensions of the general horizontal rectangle, we can express the volume $dV$ of the solid created by rotating one of these rectangles around the y-axis as $$dV=[\pi \cdot (Outer Radius)^2 - \pi \cdot (InnerRadius)^2 ] \cdot (thickness) $$$$=[\pi \cdot (y+2)^2 - \pi \cdot (y^2)^2]\cdot dy$$ $$=\pi \cdot [(y+2)^2-(y^2)^2]\cdot dy$$ $$dV=\pi \cdot (-y^4 + y^2 + 4y + 4)\cdot dy$$

All that is left is to integrate, with limits of integration $y=-1$ and $y=2$:

$$Volume = \int_{-1}^2dV$$ $$= \int_{-1}^2\pi \cdot (-y^4 + y^2 + 4y + 4)\cdot dy$$ $$= \pi [- \frac{1}{5} y^5 + \frac{1}{3} y^3 + 2y^2 + 4y]_{-1}^2$$ $$Volume = \frac{72 \pi}{5} = 14.4 \pi (units^3)$$

Answer 2

First find the limits of integration: $$y^2 =y+2 \\ y^2-y-2=0 \longrightarrow(y-2)(y+1)=0 \\ y=2 \;\lor \;y=-1$$

Then setup an equation in the form $$\int_{-1}^2 \pi\left(\left( y+2\right)^2-y^4\right) dy = \frac{72}{5}\pi$$