The solid with a semicircular base of radius 5 whose cross sections perpendicular to the base and parallel to the diameter are squares How would I go about solving this problem? I'm stuck.

For a square base parallel to the x-axis, you can use the formula: $$ \int_a^b f\left(y\right)^2 dy $$ The equation for a circle is $$ x^2+y^2=r^2 \\\ x=\sqrt{25-y^2} $$ The endpoints of integration are $y=0$ to $y=5$. Also, since rewriting this formula, we only get the half of the circle in positive in $x$, so we have to double before substituting. $$ \int_0^5 \left(2\sqrt{25-y^2}\right)^2 dy $$