Volume Semircircle Triangle 3C78F8

1. **State the problem:** We want to find the volume of a solid whose base is a right triangle with vertices at $(0,0)$, $(7,0)$, and $(0,7)$. The cross sections perpendicular to the base and parallel to the $y$-axis are semicircles. 2. **Understand the base and slicing direction:** The base is the triangle bounded by the lines $x=0$, $y=0$, and the line connecting $(7,0)$ to $(0,7)$. The equation of the hypotenuse line is found by: $$y = -x + 7$$ 3. **Set up the slicing:** Since the cross sections are perpendicular to the base and parallel to the $y$-axis, slices are taken at constant $x$ values. At each $x$, the cross section is a semicircle with diameter equal to the length of the segment in the $y$-direction inside the triangle. 4. **Find the length of the diameter at position $x$:** The base of the triangle extends from $y=0$ to $y = -x + 7$. So the diameter $d(x)$ of the semicircle is: $$d(x) = -x + 7$$ 5. **Area of the semicircle cross section:** The radius $r(x)$ is half the diameter: $$r(x) = \frac{d(x)}{2} = \frac{-x + 7}{2}$$ The area of a full circle is $\pi r^2$, so the area of a semicircle is half of that: $$A(x) = \frac{1}{2} \pi r(x)^2 = \frac{1}{2} \pi \left(\frac{-x + 7}{2}\right)^2 = \frac{\pi}{2} \frac{(-x + 7)^2}{4} = \frac{\pi}{8} (-x + 7)^2$$ 6. **Set up the integral for volume:** The volume $V$ is the integral of the cross-sectional area along $x$ from $0$ to $7$: $$V = \int_0^7 A(x) \, dx = \int_0^7 \frac{\pi}{8} (-x + 7)^2 \, dx$$ 7. **Simplify and compute the integral:** Expand the square: $$(-x + 7)^2 = (7 - x)^2 = 49 - 14x + x^2$$ So, $$V = \frac{\pi}{8} \int_0^7 (49 - 14x + x^2) \, dx$$ 8. **Integrate term-by-term:** $$\int_0^7 49 \, dx = 49x \Big|_0^7 = 49 \times 7 = 343$$ $$\int_0^7 -14x \, dx = -14 \frac{x^2}{2} \Big|_0^7 = -7 x^2 \Big|_0^7 = -7 \times 49 = -343$$ $$\int_0^7 x^2 \, dx = \frac{x^3}{3} \Big|_0^7 = \frac{343}{3}$$ 9. **Sum the integrals:** $$343 - 343 + \frac{343}{3} = 0 + \frac{343}{3} = \frac{343}{3}$$ 10. **Calculate the volume:** $$V = \frac{\pi}{8} \times \frac{343}{3} = \frac{343 \pi}{24}$$ **Final answer:** $$\boxed{\frac{343 \pi}{24}}$$