Region $R$ is enclosed by the curve $y=\frac{1}{2} x^{2}$ and the line $y=2 x$.

What is the volume of the solid generated when $R$ is rotated about the $y$ axis?
Give an exact answer in terms of $\pi$.

Question

Region $R$ is enclosed by the curve $y=\frac{1}{2} x^{2}$ and the line $y=2 x$.

What is the volume of the solid generated when $R$ is rotated about the $y$ axis?
Give an exact answer in terms of $\pi$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the volume of the solid generated when region $R$ is rotated about the $y$-axis, we use the method of cylindrical shells. The volume $V$ is given by the integral formula: $$ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx $$ where $f(x)$ is the height of the shell and $x$ is the radius. ‖ ‖ step 2 ⋮ The region $R$ is enclosed by the curves $y = \frac{1}{2}x^2$ and $y = 2x$. We need to find the points of intersection to determine the limits of integration. Setting $\frac{1}{2}x^2 = 2x$, we solve for $x$: $$ \frac{1}{2}x^2 = 2x $$ $$ x^2 = 4x $$ $$ x(x - 4) = 0 $$ Thus, $x = 0$ and $x = 4$. ‖ ‖ step 3 ⋮ The volume integral becomes: $$ V = 2\pi \int_{0}^{4} x \left(2x - \frac{1}{2}x^2\right) \, dx $$ Simplifying the integrand: $$ V = 2\pi \int_{0}^{4} (2x^2 - \frac{1}{2}x^3) \, dx $$ ‖ ‖ step 4 ⋮ We now integrate term by term: $$ V = 2\pi \left[ \int_{0}^{4} 2x^2 \, dx - \int_{0}^{4} \frac{1}{2}x^3 \, dx \right] $$ $$ V = 2\pi \left[ 2 \int_{0}^{4} x^2 \, dx - \frac{1}{2} \int_{0}^{4} x^3 \, dx \right] $$ ‖ ‖ step 5 ⋮ Compute the integrals: $$ \int_{0}^{4} x^2 \, dx = \left. \frac{x^3}{3} \right|_{0}^{4} = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} $$ $$ \int_{0}^{4} x^3 \, dx = \left. \frac{x^4}{4} \right|_{0}^{4} = \frac{4^4}{4} - \frac{0^4}{4} = 64 $$ ‖ ‖ step 6 ⋮ Substitute back into the volume formula: $$ V = 2\pi \left[ 2 \cdot \frac{64}{3} - \frac{1}{2} \cdot 64 \right] $$ $$ V = 2\pi \left[ \frac{128}{3} - 32 \right] $$ $$ V = 2\pi \left[ \frac{128}{3} - \frac{96}{3} \right] $$ $$ V = 2\pi \cdot \frac{32}{3} $$ $$ V = \frac{64\pi}{3} $$ ‖ ∻Answer∻ ⚹ The exact volume of the solid generated when region $R$ is rotated about the $y$-axis is $\frac{64\pi}{3}$. ⚹ ∻Key Concept∻ ⚹Volume of solid of revolution using cylindrical shells⚹ ∻Explanation∻ ⚹The volume is calculated by integrating the product of the radius and height of cylindrical shells formed by rotating the region around the $y$-axis.⚹◊What is the formula for finding the volume of a solid generated by rotating a region about the $y$-axis?⍭ Generate me a similar question◊

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