So I have a solid with a semicircular base of radius 19. The cross sections are perpendicular to the base and parallel to the diameter and these cross sections are squares.
The semi circle is placed on the xy plane so that the diameter is on the x axis and it is centered on the y axis.
I need help setting up the integral that will give the volume of the solid. I don't know how to go about it. I've read a bit and the only thing I can figure out is that it is bounded from 0 to 19 as the integral boundary.
$$
S : \frac{x^2}{4} + \frac{y^2}{9} = 1
$$
Because the cross sections are isosceles right triangles with hypotenuse in the base, the length of the hypotenuse is $ 6\sqrt{1-\frac{x^2}{4}}$, which means that the area of the cross section is $9(1-\frac{x^2}{4})$
Best Answer
the semi-circle is centred on the origin, so its equation is $ x^2+y^2=19^2: y \ge 0$
so for any given $y$ the integrand must be the area of the square whose sides are $2 \sqrt{19^2-y^2}$
$$V=4\int_0^{19} (19^2-y^2) dy$$