▶️ Answer/Explanation
Solution
(a) Area of R:
Method 1: Vertical slices \[ \text{Area} = \int_0^A \tan x \, dx + \int_A^{2^{1/3}} (2 – x^3) \, dx \approx 0.729 \] where A ≈ 0.902155
Method 2: Horizontal slices \[ \text{Area} = \int_0^B [(2 – y)^{1/3} – \tan^{-1}y] \, dy \approx 0.729 \] where B ≈ 1.265751
(b) Area of S:
Method 1: Vertical slices \[ \text{Area} = \int_0^A (2 – x^3 – \tan x) \, dx \approx 1.160 \]
Method 2: Horizontal slices \[ \text{Area} = \int_0^B \tan^{-1}y \, dy + \int_B^2 (2 – y)^{1/3} \, dy \approx 1.160 \]
(c) Volume of revolution:
Using washer method: \[ V = \pi \int_0^A [(2 – x^3)^2 – \tan^2 x] \, dx \approx 2.652\pi \approx 8.331 \]