The two parabolas intersect for
[tex]8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2[/tex]
and so the base of each solid is the set
[tex]B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}[/tex]
The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, [tex]|x^2-(8-x^2)| = 2|x^2-4|[/tex]. But since -2 ≤ x ≤ 2, this reduces to [tex]2(x^2-4)[/tex].
a. Square cross sections will contribute a volume of
[tex]\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x[/tex]
where ∆x is the thickness of the section. Then the volume would be
[tex]\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}[/tex]
where we take advantage of symmetry in the first line.
b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of
[tex]\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x[/tex]
We end up with the same integral as before except for the leading constant:
[tex]\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx[/tex]
Using the result of part (a), the volume is
[tex]\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}[/tex]
c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is
[tex]\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x[/tex]
and using the result of part (a) again, the volume is
[tex]\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}[/tex]
