Calculus- Shell Method Problem - find the volume of the solid when the region $y^2=x, x=0,$ and $y=1$ is rotated around $x=-3.$

Question

Since the problem uses $y^2=x$, I first assumed that the element must be horizontal (parallel to the $x$-axis). However, the bounded region has all $y$ values greater than $0$, so I could also use a vertical element. This problem has me stumped; I know how to set up the integral but for the shell method I need to find the radius (element to axis of rotation) and the height of the element.

What is the best way to approach this problem?

Answer 1

for the shell method you need the following:

(a) the radius of the cylinder. in your case you have $x + 3$

(b) thickness of the cylinder is $dx$

(c) height of the cylinder. in your case it is $1 - y$

(d) the range of the variable is $x: 0 \to 1$

infinitesimal volume of the shell is $$dV = 2 \pi (1-y)(x+3) \, dx, V = 2 \pi \int_0^1(1-y)(3 + x) \, dx, \text{ where } y = \sqrt x $$

i trust that you can evaluate this integral.

Answer 2

Disk method is best way:

$V = \displaystyle \int_{0}^1 \pi\left((y^2+3)^2- 3^2\right)dy$.

Shell method is:

$V = \displaystyle \int_{0}^1 2\pi(x+3)(1-\sqrt{x})dx$