Calculus 2 Integration of area to find a volume

Question

I am having trouble with this problem in my Calculus 2 class where I need to fine a volume by integrating an area.

In this problem, we will find the volume of a solid with circular base of radius $2,$ for which parallel cross-sections perpendicular to the base are squares. To do this, we will assume that the base is the circle $x^2+y^2=4,$ so that the solid lies between planes parallel to the $x$-axis at $x=2$ and $x=−2.$ The cross-sections perpendicular to the $x$-axis are then squares whose bases run from the semicircle $y=−\sqrt{4−x^2}$ to the semicircle $y=\sqrt{4-x^2}.$

What is the area of the cross-section at $x$?

What is the volume of the solid?

Answer 1

$$x^2 + y^2 = 4 \implies y = \pm \sqrt {4-x^2}$$

For each cross-section, the length of the edge is the range of y values.

$$s = 2\sqrt {4-x^2}$$

The area of each cross-section is $s^2$:

$$\int_{-2}^2 4(4-x^2) \ dx$$

Answer 2

At each point $x,$ first calculate the length of the base of the square, namely $$\sqrt{4-x^2}-\left(-\sqrt{4-x^2}\right)=2\sqrt{4-x^2}.$$ Then the area of each square is given by $\left(2\sqrt{4-x^2}\right)^2=4\left(4-x^2\right).$ Then you integrate $4\left(4-x^2\right)\mathrm d x$ from $-2$ to $2$ to get the required volume.