## University Calculus: Early Transcendentals (3rd Edition)

$$\dfrac{ 5\pi}{6}$$
Consider the shell model to compute the volume: $$Volume=\int_{m}^{n} (2 \pi) (\space Radius \space of \space shell) \times ( height \space \text{of} \space \text {shell}) dx \\ = (2 \pi) \times \int_{0}^{1} \cdot y (2-y-y^2) dy \\=( 2\pi) [y^2-\dfrac{y^3}{3}-\dfrac{y^4}{4}]_{0}^{1} \\=[-\dfrac{1}{3}(2 \pi)-\dfrac{1}{4}(2 \pi)] \\=\dfrac{ 5\pi}{6}$$