The base of a solid is a circle centered at the origin with a radius r, and every plane section perpendicular to a diameter is a square. What is the volume of the solid?
So this would usually be easy to solve, but they don't actually give any numbers. I think the equation of the circle here is $x^2+y^2=r$, which should equal to $y=2\sqrt{2-x^2}$, but I'm not sure if this is correct or where to go from there.
If you want to find the volume, you 'add up' all the 'infinitely small' areas - in this case squares. This image here will give you an idea of what is happening. Taking the limit of this adding process, you get an integral. So What we have is $$ \int A(x) \;dx $$ where $A(x)$ gives the area of the square at $x$. This way the integral does the 'adding' and the integrand is what is being added - the areas - to obtain the volume. Now all we need is the limits and $A(x)$. We know that the base is a circle, which as you noted is $x^2+y^2=r^2$. A side of the square will lie in the plane where the circle is. So say you are at $x$-value $x=a$. Then you know that $y= \sqrt{r^2-a^2}$. Or generally, $y= \sqrt{r^2-x^2}$. But that's only half the length of a side, so the side of the square is $2\sqrt{r^2-x^2}$. But we know $A(x)=s^2$, where $s$ is the length of a side. So $$ A(x)= s^2= \left(2\sqrt{r^2-x^2}\right)^2= 4(r^2-x^2) $$ We just need to 'add up' all the possible squares, i.e. find out limits of integration. We need to go across the possible $x$ values for the circle, which are from $-r$ to $r$. So we have $$ V= \int A(x) \;dx= \int_{-r}^r 4(r^2-x^2) \;dx $$ Which I will leave as for you as practice to integrate out.