I need a quick proof here. I've done this problem in both Shell and Disk Method, and get a different answer for each respectively. The book says the right answer is 8π and uses Disk Method, I agree, but when I use Shell Method, I get 18π. Can someone tell me my mistake please?
On
$$\int_0^3 \pi x^2\mathsf dx = \int_0^3 2 \pi y(3-y)\mathsf d y = 9\pi$$
The first method cuts the cone into discs of radius $x$ perpendicular to the x-axis; for $x\in[0;3]$. These discs have area of $\pi x^2$.
The second method cuts the cone into shells of radius $y$ and height $(3-y)$ parallel to the x-axis; for $y\in[0;3]$. The cylindric shells are generated by rotating the line segments $\overline{(y,y)(3,y)}$ in circular orbits around the x-axis. These have length $3-y$, and trace circles of radius $y$, and thus generate cylindric shells of surface area $2\pi y(3-y)$.
For the shell method, it should be $$2\pi\int_{0}^3(y)(3-y)\,dy$$ If you draw a horizontal strip at level $y$, you get$\;r(y)=y,\;h(y)=3-y$.