Find volume of solid when rotating on x-axis and find volume of solid when rotating on y-axis with the same equation.

Question

The given values are $y = 2x^2$, $y=0$ , $x=0$ , $x=5$.

I need to find the volume of solid when rotating on the x-axis and also the volume of solid when rotating on the y-axis using the same equation.

I'm pretty convinced I solved the x-axis portion correctly using the disc method.

$$V_x = \pi\int_0^5 (2x^2)^2 dx = 2500\pi$$

I'm totally lost on how to solve the equation if it's rotating on the y-axis.

My first guess is the washer method. That leaves me at this though.

$$V_y = \pi\int_0^{50} (\sqrt\frac{y}{2})^2 - \int_0^{50} ?^2 $$

Assuming I'm on the right track, what's my 2nd value that I'll use to minus?

Also is 50 the correct upper boundary or still 5?

Answer 1

Use:

• Rotating around the x-axis: $$\text{V}_x=\pi\int_a^b f(x)^2\space\text{d}x$$
• Rotating around the y-axis: $$\text{V}_y=2\pi\int_a^b xf(x)\space\text{d}x$$

So,we will get:

1. $$\text{V}_x=\pi\int_0^5 (2x^2)^2\space\text{d}x=4\pi\int_0^5 x^4\space\text{d}x=2500\pi$$
2. $$\text{V}_y=2\pi\int_0^5 x(2x^2)\space\text{d}x=4\pi\int_0^5 x^3\space\text{d}x=625\pi$$