Is the volume of rotating $y=x^2$ around the y-axis equal to the volume of rotating half of the function $(x\geq0)$ around the y-axis?

Question

We know that the volume of the shape formed by rotating $x=\sqrt{y}$ around the y-axis is $$\pi\int_{0}^{\infty} y dy$$ Is this equal to the volume of the shape formed by turning $y=x^2$ around the y-axis at $x=0$?

Answer 1

You should be cautious of $\infty$ as the upper bound because the integral here diverges. If the upper bound is finite, then, yes, the two volumes are equal.