Question

VolumesFind the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³  and the lines x = 1 and y = 1/2 abouta. the x-axis

Accepted Answer

First, identify the region bounded by the curve $$y = \frac{4}{x^3$}$$, the vertical line $$x = 1$$, and the horizontal line $$y = \frac{1}{2}. To $$find the limits of integration, determine the $$x-$$values where $$y = \frac{1}{2}$$ intersects the curve by solving $$\frac{4}{x^3$} = \frac{1}{2}. $$Solve for $$x in $$the equation $$\frac{4}{x^3$} = \frac{1}{2} to $$find the lower limit of integration. This will give you the interval $$[a, 1]$$ over which the region is bounded, where $$a is $$the solution to the equation. Set up the volume integral using the disk method since the solid is generated by revolving the region around the x-axis. The volume $$V is $$given by $$V = \pi \int_a^1$ [f(x)]^2 \, dx$$, where $$f(x) is $$the function representing the radius of the disks, which in this case is $$y = \frac{4}{x^3$}. $$Write the integral explicitly as $$V = \pi \int_a^1$ \left( \frac{4}{x^3$} \$$right)^2$$ \, dx = \pi \int_a^1$ \frac{16}{x^6$} \, dx. $$This integral represents the volume of the solid formed by revolving the region around the x-axis. Evaluate the integral $$\int_a^1$ \frac{16}{x^6$} \, dx by $$applying the power rule for integration, then multiply the result by $$\pi to $$express the volume. Remember to substitute the limits $$a$$ and $$1$$ after integrating.

Volumes
Find the volume of the solid generated by revolving the “triangular” region bounded by the curve y = 4/x³ and the lines x = 1 and y = 1/2 about
a. the x-axis

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Key Concepts

Volume of Solids of Revolution

Disk/Washer Method

Setting Up Integration Limits and Boundaries