# Find the volume of the solid formed by the revolving the region around a line

I have to find the volume of the solid formed by the revolving the region enclosed by $x=\frac{y^2}4$ and $y=x^5+x^3$ around the line $y=2$

I know how to find the volume when it revolve around x-axis, but when it revolve around a function line or y-axis, how should I approach it? Appreciate anyone help

## 2 Answers

Transform $y$ into $y-2$, so that line $y=2$ is transformed into the $x$-axis. The two equations become $y=x^5+x^3-2$ and $y=2\sqrt x -2$.

• I got it, thanks – user3676506 Dec 22 '15 at 22:06 Intersection of the planes that are perpendicular to the x-axis and the solid is the region(washer) with the area $$\pi[(2-x^5-x^3)^2-(2-2 \sqrt x)^2]$$. Therefore, $$V= \int_0^1 \pi[(2-x^5-x^3)^2-(2-2 \sqrt x)^2]dx$$

• Is there a way to know which function minus which without drawing a graph? – user3676506 Dec 22 '15 at 22:44
• You can make a sign chart to see that, here I drew a graph to see the washer. – frosh Dec 22 '15 at 22:47
• And, I used the general definition of volume, i.e., $\int_a^bA(x)dx$ where $A(x)$ is the cross section area function. Perhaps you enjoy reading the Cavalieri's Principle which relies on this definition: en.wikipedia.org/wiki/Cavalieri%27s_principle. – frosh Dec 22 '15 at 22:50