# The solid with a semicircular base of radius 5 whose cross sections perpendicular to the base and parallel to the diameter are squares [closed]

How would I go about solving this problem? I'm stuck.

• Please edit the body of the Question so that it is self-contained. The burden of posing the problem should not be borne exclusively by the title. Putting the circumstances together with the question asked gives you an opportunity to explain what is understood and what is giving difficulty, so that Readers have a better idea how to respond. – hardmath Sep 11 '15 at 14:52
• Possible duplicate math.stackexchange.com/questions/1339184 – John Joy Sep 11 '15 at 15:09

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. $$\int_0^5 \left(2\sqrt{25-y^2}\right)^2 dy$$