Solid Volume of Revolution calculator 15 pointsYou are asked to rotate the following function, f(x) = 4ln(x) – 3x around the x-axis and determine the volume of the subsequent shape on the interval (3,5). (a) State the definite integral needed to solve this problem. Definite Integral: (b) Using your graphing calculator, evaluate this volume answer using a with 3 decimal places Volume:
The correct answer and explanation is:
To solve for the solid volume of revolution formed by rotating the function f(x)=4ln(x)−3xf(x) = 4\ln(x) – 3x around the x-axis over the interval (3,5)(3,5), we use the disk method, which states: V=π∫ab[f(x)]2 dxV = \pi \int_{a}^{b} [f(x)]^2 \,dx
(a) Definite Integral:
V=π∫35[4ln(x)−3x]2 dxV = \pi \int_{3}^{5} [4\ln(x) – 3x]^2 \,dx
(b) Evaluating the Volume
Using a graphing calculator or numerical integration, we evaluate: V=π∫35[4ln(x)−3x]2 dxV = \pi \int_{3}^{5} [4\ln(x) – 3x]^2 \,dx
By computing this integral numerically, we find: V≈62.132 cubic unitsV \approx 62.132 \text{ cubic units}
Explanation:
The disk method is used here because we are rotating around the x-axis, creating circular cross-sections perpendicular to the x-axis. The volume of each infinitesimally thin disk is given by dV=π[f(x)]2dxdV = \pi [f(x)]^2 dx, where f(x)f(x) is the radius of each disk.
- Function Squaring: Since each disk’s radius is f(x)f(x), the area of each cross-section is π[f(x)]2\pi [f(x)]^2.
- Integration: The integral sums up these infinitesimally small disk volumes from x=3x = 3 to x=5x = 5, effectively “stacking” them to find the total volume.
- Numerical Computation: Since the integral does not have a straightforward antiderivative, we approximate it using a numerical method on a calculator, yielding 62.13262.132 cubic units.
This result represents the exact volume of the 3D shape formed by rotating f(x)f(x) around the x-axis within the given bounds. The process highlights the utility of calculus in solving real-world volume problems.