Quick Area of Cross Section Question &quot;The base of a solid is the region in the first quadrant between the graph of $y=2x$ and the $x$-axis for $0\leq1\leq2$. For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the $x$-axis and one radius in the $xy$-plane. What is the volume of the solid?&quot;
I guess I'm a bit confused about both the wording and where I'd have to go in this question, I understand how to find the area of a cross-section when these cross-sections might be a square or a rectangle, but I'm a bit lost when it comes to circles as well. Any help would be appreciated!
THe method of using integraion to find find volume that you are using is

Pick an axis.
Slice the volume in slices perpendicular to that axis.
Integrate the volume of the slices with respect to the coordinate along that axis.

(Generally, the axis need not be a coordinate axis, but it is convenient when this happens.)
When the problem has

For the solid, each cross section perpendicular to the $x$-axis is a quarter circle with the corresponding circle's center on the x-axis and one radius in the $xy$-plane.

You are told that the slice is a quarter circle, so has one quarter the area of a circle, and that (1) all the radii for that quarter circular are perpendicular to the $x$-axis and (2) one radius is in the $xy$-plane.  This means that in a slice, one can take the distance perpendicular to the $x$-axis to the line $y = 2x$ as the radius of the circle, one quarter of which area is accumulated into the integral.  (In the $xy$-plane, a perpendicular to the $x$-axis is parallel to the $y$-axis.  A ray starting on the $x$ axis, in the first quadrant of the $xy$-plane, and proceeding perpendicular to the axis crosses the triangle base of the volume, reaching the end of the base at the line $y = 2x$.  (If this is not clear, sketch a graph of the base.)
Suppose $r$ is the radius of a circle.  The area of that circle is $\pi r^2$.  The area of one-quarter of that circle is $\frac{1}{4} \pi r^2$.  The distance from the $x$-axis to the line $y = 2x$ (the radius described by the ray in the previous paragraph) is $2x$, so the radius we are to take for the slice at coordinate $x$ is $r = 2x$.  Then the area contribution from the slice at coordinate $x$ is $\frac{1}{4} \pi (2x)^2$.  That is the integrand that we integrate from $x = 0$ to $x = 2$.  (Again, if it is not clear that this is the range of coordinate that includes all the slices, sketch the base.)
The following diagram of the volume is from a point in the $(+++)$-octant looking back throught the volume at the origin.

Example slices (for $x \in \{1/2, 1, 3/2, 2\}$ are shown as fairly clear quarter circles.  The base is the $xy$-plane, at the bottom of the diagram.  The dashed gray line is in the $xz$-plane and marks the ends of the radii of the quarter circles in that plane.
We can more clearly see the &quot;slicing&quot; viewing from the $(+-+)$-octant nearly parallel to the positive $y$-axis.

This angle reinforces that we are given the axis, we are given the area formula for the slices, and all we have to find is the relation between the $x$-coordinate and the area of the corresponding slice.
The hardest part of this problem is understanding the wording. Trying to describe a $3$-d shape using words is sometimes difficult, but what it is trying to describe is a pretty simple shape. The base is a right triangle with a length of two units and a height of $4$ units. To find the total volume we need to find how this shape looks in the $z$-direction. If you take a cross-section perpendicular to the $x$-axis you should see a quarter circle, with radius $2x$, because that is the distance from the $x$-axis to the given line. from this we can tell that one cross section has volume $(\pi/4) \cdot (2x)^2 \, dx$. This can be simplified to the $\pi \cdot x^2 \, dy$. Now we simply take the integral from $0$ to $2$ of this volume of a single cross section to add up all the cross sections from $x=0$ to $x=2$. This equation looks like the following:
$$ \pi \, \int_{0}^{2} x^2 \, dx = \frac{8 \, \pi}{3}.$$