The base of a solid is a region in the first quadrant bounded by the x-axis, the y-axis, and he line x+2y=8. If the cross sections of the solid perpendicular to the x-axis are semicircles, what is the volume of the solid?

Question

Samantha Adkison · Accepted Answer

See the answer below:

Scarlett Allison · Answer

undefined To find the volume of the solid, you integrate the area of each cross-section perpendicular to the x-axis from the starting x-value to the ending x-value.

The equation of the line is $x + 2y = 8$, rearranging to $y = \frac{8 - x}{2}$. This line intersects the x-axis at $x = 8$ and the y-axis at $y = 4$.

The region bounded by the x-axis, the y-axis, and the line $x + 2y = 8$ is a triangle with base 8 and height 4.

The area of each cross-section perpendicular to the x-axis is a semicircle, and the radius of each semicircle is the value of y. Thus, the area of each cross-section is $\frac{\pi y^2}{2}$.

Integrating from $x = 0$ to $x = 8$, the volume $V$ of the solid is given by:

$$V = \int_{0}^{8} \frac{\pi y^2}{2} dx$$

Substituting $y = \frac{8 - x}{2}$ into the integral expression:

$$V = \int_{0}^{8} \frac{\pi (\frac{8 - x}{2})^2}{2} dx$$

$$V = \int_{0}^{8} \frac{\pi (64 - 32x + x^2)}{8} dx$$

$$V = \frac{\pi}{8} \int_{0}^{8} (64 - 32x + x^2) dx$$

$$V = \frac{\pi}{8} \left[ 64x - 16x^2 + \frac{x^3}{3} \right]_{0}^{8}$$

$$V = \frac{\pi}{8} \left[ (64 \times 8 - 16 \times 8^2 + \frac{8^3}{3}) - (0 - 0 + 0) \right]$$

$$V = \frac{\pi}{8} \left[ (512 - 1024 + \frac{512}{3}) \right]$$

$$V = \frac{\pi}{8} \left[ \frac{512}{3} \right]$$

$$V = \frac{\pi}{3} \times 16$$

$$V = \frac{16\pi}{3}$$

So, the volume of the solid is $\frac{16\pi}{3}$ cubic units.