Question

The region R is bounded by the graph of f(x)=2x(2−x) and the x-axis. Which is greater, the volume of the solid generated when R is revolved about the line y=2 or the volume of the solid generated when R is revolved about the line y=0? Use integration to justify your answer.

Accepted Answer

Step 1: Identify the region R. The region R is bounded by the graph of f(x) = 2x(2 − x) and the x-axis. To find the bounds of integration, solve f(x) = 0. This gives x = 0 and x = 2 as the limits of integration. Step 2: Set up the volume integral for the solid generated when R is revolved about the line y = 2. The distance from the line y = 2 to the curve is (2 − f(x)), and the distance from the line y = 2 to the x-axis is 2. Use the washer method to compute the volume: $$ V = \pi \int_{0}^{2} \left[ (2)^2 - (2 - f(x))^2 \right] dx . $$Step 3: Expand and simplify the integrand for the volume about y = 2. Substitute f(x) = 2x(2 − x) into the expression and simplify $$ (2 - f(x))^2 . $$This will involve expanding $$ (2 - 2x(2 − x))^2 $$ and combining terms. Step 4: Set up the volume integral for the solid generated when R is revolved about the line y = 0. The distance from the line y = 0 to the curve is f(x). Use the disk method to compute the volume: $$ V = \pi \int_{0}^{2} [f(x)]^2 dx . $$Substitute f(x) = 2x(2 − x) and simplify $$ [f(x)]^2 . $$Step 5: Compare the two volumes. Evaluate both integrals (without calculating the final numerical values here) and analyze the expressions to determine which volume is greater. The comparison depends on the relative contributions of the terms in the integrals for the two setups.

The region R is bounded by the graph of f(x)=2x(2−x) and the x-axis. Which is greater, the volume of the solid generated when R is revolved about the line y=2 or the volume of the solid generated when R is revolved about the line y=0? Use integration to justify your answer.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Volume of Revolution

Definite Integrals

Comparison of Volumes