Volume obtained by rotation

Question

So, I want to calculate the volume of the area limited by these functions when it rotates around $y'Oy$. I started by $∫[0,1](πx^2)dy$ but when I wanted to get x^2, it turned out that it had $y$ in it. How do I integrate? Can I assume that $y$ is a constant? $y=sqrt(x)-x$ --> $x^2=y^2-x-2xsqrt(x)$

Answer 1

As you are integrating with respect to $y$, $\int dy$, the $y$ terms will follow the standard rules of integration, in this case; $\int y^2\ dy =\frac{y^3}{3}$.
For the $x$ terms, it is simpler as they are treated as constants in $dy$; $\int x\ dy =xy$.

Hence;

$\begin{align*}\pi\int_{0}^{1} y^2-x-2x\sqrt{x}\ dy &=\pi[\frac{y^3}{3}-xy-2x^{\frac{3}{2}}y]_{0}^{1}\\ &= \pi((\frac{1^3}{3}-x-2x^{\frac{3}{2}})-(\frac{0^3}{3}-0x-0x^{\frac{3}{2}})) \\ &=\frac{\pi}{3}-\pi{x}-2\pi{x}^{\frac{3}{2}} \end{align*}$