Question

How do you find the volume of the solid obtained by rotating the curve about the y-axis over [0,1]?

y=x`sqrt(1-x^2)`

Answer 1

We must imagine that the graph has radius `r`. Then the volume will be `pir^2h`. In this case we have:

`V = pi int_0^1 (xsqrt(1 - x^2))^2dx`

Therefore

`V = pi int_0^1 x^2(1 - x^2)dx`

`V= pi int_0^1 x^2 - x^4 dx`

`V = pi[1/3x^3 - 1/5x^5]_0^1`

`V = pi(1/3 - 1/5)`

`V = 2/15pi` cubic units

Hopefully this helps!