Answer :
Answer:
[tex]V = \frac{5\pi}{6}[/tex] or 2.62
Step-by-step explanation:
Since our region (on the left) is bounded by x = 1 and x = 0 (where [tex]y = x^2 = 0[/tex], if we take center at x = 2 then our radius will range from 1 to 2 (x=1 to x = 0). We can use the following integration to calculate the volume using shell method
[tex]V = \int\limits^2_1 {2\pi r h} \, dr[/tex]
where r = 2 - x so x = 2 - r and [tex]h = y = x^2 = (2-r)^2[/tex] for [tex]1 \leq r \leq 2[/tex]
[tex]V = \int\limits^2_1 {2\pi r(2-r)^2} \, dr[/tex]
[tex]V = \int\limits^2_1 {2\pi r(4 - 4r + r^2)} \, dr[/tex]
[tex]V = 2\pi \int\limits^2_1 {r^3 - 4r^2 +4r} \, dr[/tex]
[tex]V = 2\pi\left[\frac{r^4}{4} - \frac{4r^3}{3} + 2r^2\right]^2_1[/tex]
[tex]V = 2\pi\left[\left(\frac{2^4}{4} - \frac{4*2^3}{3} + 2*2^2\right) - \left(\frac{1^4}{4} - \frac{4*1^3}{3} + 2*1^2\right)\right][/tex]
[tex]V = 2\pi(4 - 32/3 +8 - 1/4 + 4/3 - 2)[/tex]
[tex]V = \pi(20 - 56/3 - 1/2)[/tex]
[tex]V = \pi\frac{120 - 112 - 3}{6}[/tex]
[tex]V = \frac{5\pi}{6}[/tex] or 2.62