Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
$$y=9x−(3x)^2, y=0;$$
about the y-axis.
How do I solve this when one of the y value is $0$?
Find the volume of the solid obtained by rotating the region
calculusdefinite integralsfunctions
Best Answer
Using the "washer method":
Trying to solve $y=9x-(3x)^2$ for $x$ we don't get a function, so it needs to be split into the two functions $$ x=0.5 \pm \sqrt{\frac{2.25-y}{9}}, \ y\geq 0 $$ The volume is then given by $$ V = \pi \int_{0}^{\frac{9}{4}} \left( 0.5+\sqrt{\frac{2.25-y}{9}} \right)^2 \ dy \ - \pi \int_{0}^{\frac{9}{4}} \left( 0.5-\sqrt{\frac{2.25-y}{9}} \right)^2 \ dy $$ where $y=\frac{9}{4}$ is the $y$ value where the two functions meet. This works because we are subtracting the volume from rotating the inner function from the volume of rotating the outer function (here are the two functions f(y) in Desmos):