Volume of solid of revolution around a line

Question

What is the volume of the solid that results when the region bounded by the graphs of $y=\sqrt{x}$, y = 0, and x=4 is revolved about the line x = 4.

My solution is $$ \pi \int_0^2 (4-y^2)^2 \, dy = \frac{256\pi}{15} $$

I don't know if this correct. How can I integrate this in terms of x? In terms of x, is the result the same with my integration above?

Answer 1

In terms of $x$, you can write as below -

$V = \displaystyle 2\pi \int_0^4 (4-x)\sqrt x \, dx$

The horizontal area of the circumference of thickness $dx$ is $2 \pi (4-x) dx \,$ where $(4-x)$ is the radius of the circumference. $y = (\sqrt x - 0)$ is the height.

Answer 2

Integrate over x with the variable change $y=\sqrt{x}$

$$ \pi \int_0^2 (4-y^2)^2 \, dy = \pi \int_0^4 (4-x)^2 \frac{dx}{2\sqrt{x}}\\ = \pi \int_0^4 \left(\frac{8}{\sqrt{x}}- 4{\sqrt{x}} + \frac12x\sqrt{x}\right)dx= \frac{256\pi}{15} $$