Consider the region bounded by $y^2=x^3$, the $x$ axis and $x=4$. I want to calculate the volume of the solid of revolution of the region around $y=8$ and I can use the washer method, as the outer radius is $8-0=8$ and the inner radius is $8-x^\frac{3}{2}$. So let's calculate:
$$V_1=\pi\int_0^4 [8^2-(8-x^\frac{3}{2})^2]dx=\frac{704}{5}\pi u^3.$$ but then I'm struggling and confused as I get this result from Wolfram Alpha:
$$V_2=∫_0^4 2 π (4-x) (8-x^\frac{3}{2})dx=\frac{3456}{35}\pi u^3.$$
Why do I get a different result?
And I have another related question. I was wandering if I could get a nice plot (visualization) of the same solid of revolution. As Wolfram Alpha is giving a different result, I can't trust the plot given by the computation linked above and also showed below.
Thank you!
The integral $V_2$ is for the volume when we rotate the region below $y=8$ and above $y=x^{3/2}$ about the vertical line $x=4$. It uses the method of cylindrical shells. So note that (i) the region is different and (ii) the axis of rotation is different.
You noted in a comment that when we rotate the region between $y=x^{3/2}$ and $x=0$ about $x=4$, we get volume $\frac{1024}{35}\pi$. When we add the volume cpmputed by Alpha, the sum simplifies to $128\pi$, the volume of a cylinder radius $4$ and base $8$.
The integral $V_1$ computes the volume when we rotate the region below $y=x^{3/2}$ and above the $x$-axis about the line $y=8$. This is the region and axis of rotation specified in the problem.