Answer to Question #131420 in Calculus for Moel Tariburu

Question

The base of a certain solid is the region enclosed by the semicircle

Answer 1

$$Solution$$

Lets say $x^2+y^2=9, for\ 0 \le x\le 3.$ We calculate the volume of the circle as;

$x^2+y^2=9\\ for\ 0 \le x \le 3\\ y^2=g-x^2\\ y=\sqrt{g-x^2}= radius\ of \ semicircles\\ so,\ area(A)=\frac{1}{2}\pi r^2 \implies \frac{1}{2}\pi (\sqrt{g-x^2})^2\\ so,\ volume(V)=\int_{x_1}^{x_2} A(x) \cdot dx=\int_{0}^{3} \frac{1}{2}\pi (\sqrt{g-x^2})^2 \cdot dx\\ =\frac{\pi}{2}[gx-\frac{x^3}{3}]_{0}^{3}=\frac{\pi}{2}[g(3-0)-\frac{1}{3}{3}(3^3-0)]\\ =\frac{\pi}{2}[27-9]=\frac{\pi}{2} \cdot 18=g \pi\ (units)^3--------->Answer$