Find the volume of the solid obtained by rotating the region in the first quadrant

Question

bounded by $$y=x^{4}, y=1,$$ and the y-axis around the y-axis.

So graphed these two points.

Intersection is at $(1,1)$

So the volume should be the definite integral between $0$ and $1$ of $$\pi(1-x^4)$$ This comes to $\dfrac{4\pi}{5}$ which is apparently not the right answer. Where am I going wrong here?

Answer 1

What about using cylindrical coordinates to calculate the volume bounded by the unit disk and $z=r^4$, then subtracting this from the volume of the solid bounded by the unit disk and $z=1$? The volume of the former (remembering $dx\,dy=r\,dr\,d\theta$) is $$\int_0^{2\pi}\int_0^1 r^4\cdot r\,dr\,d\theta=\frac{\pi}{3}.$$ The volume of the latter is $\pi$, so the volume we want is $$\pi-\frac{\pi}{3}=\frac{2\pi}{3}.$$

Answer 2

HINT...the volume when rotated about the y axis is $$\pi\int x^2 dy$$ therefore you need to evaluate $$\pi\int_0^1x^2\frac{dy}{dx}dx$$