The area under $y=3x-x^2$ between $x\in[0,2]$ is rotated about the $y$-axis and I need to find the volume of the generated solid.
Using the cylindrical shell method, I found the volume to be $8\pi$.
Using the disc method, I get a washer and the integral becomes $$ v=3\pi\int\sqrt{9-4y}\,\mathrm{d}y $$ However, I am not sure what the bounds of the integral is supposed to be. If I use
$$ v=3\pi\int_0^{2.25}\sqrt{9-4y}\,\mathrm{d}y $$ Then I get the volume to be $\frac{27}{2}\pi$ which I guess is wrong and does not agree with the answer I got from cylindrical shell method. What am I doing wrong in my disc method? What the bounds of the integrals are supposed to be?
First we ought to split into cases depending on whether $y$ is bigger than or smaller than $2$. For $y$ below $2$, the outer radius of the washer is $2$, while for $y>2$, the outer edge of the washer is given by the graph.
The inner radius of the washer is in either case $$ \frac{3-\sqrt{9 - 4y}}{2} $$ while for the upper part, the outer radius is given by $$ \frac{3+\sqrt{9-4y}}{2} $$ The volume is thus $$ V = \int_0^{2.25}A(y)\,dy\\ = \int_0^2\pi\left(2^2-\left(\frac{3-\sqrt{9-4y}}{2}\right)^2\right)\,dy\\ + \int_2^{2.25}\pi\left(\left(\frac{3+\sqrt{9-4y}}{2}\right)^2 - \left(\frac{3-\sqrt{9-4y}}{2}\right)^2\right)\,dy\\ = \int_0^2\pi\left(y-\frac12+\frac32\sqrt{9-4y}\right)dy + \int_2^{2.25}3\pi\sqrt{9-4y}\,dy $$