- #1
- 1,120
- 1
Hi could you please say if this is right or wrong and if it is not where I went wrong.
I got the question:
Find the volume of the area between y=x-x^3 from 0 to 1 revolved about the y-axis.
So here are my workings:
[tex]V = \pi \int_0^1{x^2dy}[/tex]
[tex]V = \pi \int_0^1{x^2 \frac{dy}{dx}dx}[/tex]
[tex]\frac{dy}{dx} = 1 - 3x^2[/tex]
[tex]V = \pi \int_0^1{x^2 \left(1 - 3x^2 \right)dx} [/tex]
[tex]V = \pi \int_0^1{x^2 - 3x^4 dx}[/tex]
[tex]V = \pi \left[ \frac{x^3}{3} - \frac{3x^5}{5} \right]_0^1 [/tex]
[tex]V = \pi \left \left( \frac{1}{3} - \frac{3}{5} \right) - \pi(0)[/tex]
[tex]V = \frac{-4\pi}{15} [/tex]
I got the question:
Find the volume of the area between y=x-x^3 from 0 to 1 revolved about the y-axis.
So here are my workings:
[tex]V = \pi \int_0^1{x^2dy}[/tex]
[tex]V = \pi \int_0^1{x^2 \frac{dy}{dx}dx}[/tex]
[tex]\frac{dy}{dx} = 1 - 3x^2[/tex]
[tex]V = \pi \int_0^1{x^2 \left(1 - 3x^2 \right)dx} [/tex]
[tex]V = \pi \int_0^1{x^2 - 3x^4 dx}[/tex]
[tex]V = \pi \left[ \frac{x^3}{3} - \frac{3x^5}{5} \right]_0^1 [/tex]
[tex]V = \pi \left \left( \frac{1}{3} - \frac{3}{5} \right) - \pi(0)[/tex]
[tex]V = \frac{-4\pi}{15} [/tex]
Last edited: