I want to compute the volume of the solid of revolution obtained by rotating the area between $f(x)=x^2+4$, $g(x)=8$ around the vertical axis $x=5$. I want to solve this problem by the 2 methods: disc method and cylinder method.
My attempt:
Disc method: If we rotate the graph $90º$ to the right then this would be equivalent to finding $2$ times the volume $V$ of the solid of revolution obtained by rotating the region under the graph of $f(x)=\sqrt{x-4}$ about the horizontal axis $y=-5$ over the interval $[4,8]$. Therefore, since $$ V=\pi \int_{4}^{8}{ \lbrace (f(x)+5)^2-5^2 \rbrace }dx=\frac{184}{3} \pi $$ the volume we are looking for is $2V=\frac{368 \pi}{3}$.
Cylinder method: Since the intersection points of the graphs occur at $x=\pm 2$ and $f(x) \leq g(x)$ on $[-2,2]$, and we want to rotate about the axis $x=5$, the cylinder method would give us $$ V=2 \pi \int_{-2}^{2}{(5-x)(g(x)-f(x))}= 2 \pi \int_{-2}^{2}{(5-x)(4-x^2)}dx=\frac{320}{3} \pi $$ As you can see, I made a mistake somewhere since the results are different, however I don't see where my mistake is.
Any help?
In advance thank you.
You're doing quite a bit of translating and it could be hard to explain, but long story short, the volume equation you have for disks:
$$V=\pi\int_{4}^{8}{((\sqrt{x-4}+5)^2-5^2)}dx={184\pi\over3}$$
represents only the left side of the parabola being rotated about $x=5$.
For the right parabola, you will need:
$$V=\pi\int_{4}^{8}{(5^2-(\sqrt{x-4}-5)^2)}dx={136\pi\over3}$$
Adding these gives ${184\pi\over3}+{136\pi\over3}={320\pi\over3}$, matching your cylindrical result, which is the correct answer.