Answer
a) See image
b) \( \int _{1}^{9}\pi(y-1)^{\frac{2}{3}}~dy\)
c) \( \frac{96\pi }{5}\)
Work Step by Step
a) See image
b) it's about the y-axis, so we need to switch \(y = x^3+1, to, x =...\)
\(x^3 = y-1,\)
\(x = \sqrt[3]{y-1}\) apply this to formula
\(volume = \int\pi x^2\), \( \int _{1}^{9}\pi(y-1)^{\frac{2}{3}}~dy\)
c) u = y-1, du = dy when y = 1, u = 0 when y = 9, u = 8.
\(\int _{0}^{8}\pi u^{\frac{2}{3}}~du\),
\(\int_{0}^{8}\pi u^{\frac{2}{3}}~du = \pi \left[ \frac{3}{5}u^{\frac{5}{3}} \right]_{0}^{8} \),
\(= \pi \left[ \frac{3}{5}(8)^{\frac{5}{3}} - \frac{3}{5}(0)^{\frac{5}{3}} \right] \)
\(= \pi \left[ \frac{3}{5}(32) \right] \)
\( = \pi \left[ \frac{96}{5} \right] \)
\(= \frac{96\pi}{5}\)
so the answer is \( \frac{96\pi}{5}\)
