The inner surface of a radially symmetric coffee mug is described as follows: consider the portion of the curve $z=x^4$ in the $x$-$z$ plane that is below the horizontal line $z=1$; then rotate it about the $z$-axis to obtain the surface.
Show that the volume of coffee mug is $\frac{2\pi}{3}$.
I am not able to see what will be the limit of the integral that I need to calculate. Please help.
Imagine to cut the mug into slice of circle disc perpendicular to z-axis. So the radius of each circle disc is x. Area is $\pi x^2$. The range of z for the mug is from z=0 to z=1. This is the limit you mentioned.
Then you can take integral.
volume = $\int^1_0 \pi x^2 dz$
Put in the equation $z=x^4$.
volume = $\int^1_0 \pi \sqrt{z} dz$.
And you can finish the rest and get the $\frac{2}{3} \pi$.