Question

Find the volume of the solid obtained by rotating the region enclosed by the curves y=sqrt of x and y = x about the y-axis.

Answer 1

Using the washer method, each washer is dy high and we need to set the equations in terms of f(y). We also need the intersection points. This can be gotten by setting both value equal.
\[x=\sqrt{x}\]
Divide each side by sqrt x.
\[\sqrt{x}=1\]
Square both sides:
\[x=1\]
Intersection is at x=1. Solve equations for f(y):
\[x=y^2\]
\[x=y\]
Plugging in x at 1 we find that y=1. So our domain is (0,1)
Formula for washer method is:
\[\pi \int\limits_{0}^{1} (OuterRadius)^2-(InnerRadius)^2\]
The inner radius in this case is x=y^2 as seen here: http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E%281%2F2%29+and+y%3Dx
Plugging in the values we get:
\[\pi \int\limits_{0}^{1} ((y)^2-(y^2)^2)dy\]
After integrating we get:
\[\pi [\frac{1}{3}y^3-\frac{1}{5}y^5]_{0}^{1}\]
Solving this we get:
\[\pi(\frac{1}{3}-\frac{1}{5})=\pi(\frac{5}{15}-\frac{3}{15})=\frac{2}{15}\pi\]