Disk and washer method for calculating the volume

Question

Is my calculation for the area enclosed by $y=x^2,y=0,x=-1,x=2$ correct?

To calculate the volume of the shape formed by rotating the area enclosed by function $y=x^2$ and $y = 0, x = 0, x=2$ around the axis $x=2$, we can use this integral $$\pi\int_{0}^{4} ydy$$.

To calculate the volume of the shape formed by rotating the area enclosed by function $y=x^2$ and $y=0, x=0, x = -1$ around the axis $x=2$, we can use this integral $$\pi\int_{0}^{1} 9-(2+\sqrt{y})^2dy$$.

Answer 1

If you want area, then: $$\int_{-1}^{2}\int_{0}^{x2}dxdy$$

Answer 2

Notice, there is mistake only in first part. Your second part is correct.

For the first part, the volume of solid generated by revolving region bounded by $y=x^2$, $y=0$, $x=0$ & $x=2$ around the line $x=2$ is given as $$\int_{0}^{4}\pi (2-x)^2dy=\pi\int_{0}^{4}(2-\sqrt y)^2dy$$

For the second part, the volume of bounded region revolved around the line $x=2$ is given as $$\int_{0}^{1}\pi (3^2-(2+\sqrt y)^2)dy=\pi\int_{0}^{1}(9-(2+\sqrt y)^2)dy$$