Volume of solid of revolution with Disk and Tube method for bounded region I have a region bounded by y axis and the curves of equation $y=\sqrt{x}$ and $y=(x-2)^2$ I need to evaluate the volume of the solid of revolution with integral but with some restriction. First, I need to evaluate when rotating this region around the x axis using the **DISK** method. Second, I need to evaluate when rotating around the line $x=2$ using the **CYLINDER** method. I'd like to understand how to determinate the integral for both method, not the answer just a start... I know that for the disk method You need to get the volume of a disk $\pi R^2E$ But I'm not sure how to define the integral considering the rotation of the region. As for the tube method, I know the equation is $2 \pi xy \delta x $ but I can't go further...

Draw a reasonably careful picture. The relevant point of intersection of the two curves is at $x=1$. The region we are rotating is crudely speaking triangular, one corner the origin, then the next corner $(1,1)$, and the final corner at $(0,4)$. The left side of the region is the straight line segment from $(0,0)$ to $(0,4)$, Now that we have the region firmly in hand, the rest is not difficult.

**First:** The curve "above" here is $y=(x-2)^2$, the curve below is $y=\sqrt{x}$. If we take a slice perpendicular to the $x$-axis at $x$, we get a cross-section which is a circle of radius $(x-2)^2$, with a circular hole of radius $\sqrt{x}$. I will leave it at that, since you wanted a start and not a full solution.

**Second:** The radius of the shell "at" $x$ is $2-x$. The height of the cylinder (tube) is $(x-2)^2-\sqrt{x}$.