Volume of a shape The base is the semicircle $$y=\sqrt{9−x^2},$$ where $x\in[-3;3]$. The cross sections perpendicular to the $x$-axis are squares.
 A: The below plot shows what @Ross notes you. In fact you have to do the following integral $$V=\int_{-3}^3(\sqrt{9-x^2})^2dx$$

A: Hint:  The idea, just like in your other question here is to chop the shape up into volumes we can simply compute and add them up.  We can easily compute the volume of a cuboid.  If we know the base is a square with side $\sqrt {9-x^2}$ and height $dx$ the volume is what?  Now integrate from $-3$ to $3$
A: First, relax and take a deep breath. Panic is counterproductive. Second, realize that you can understand these problems. Thousands of students have, and so can you. Here's a walkthrough:


*

*Can you envision what the base of your solid is? $y=\sqrt{9-x^2}$ describes a semicircle of radius 3, lying above the $x$-axis. Imagine this drawn on the floor, or actually get down and draw it on the floor. In the $x$ direction, it runs from $x=-3$ to $x=3$, as indicated in Babak's picture.

*The problem says that the cross sections of the solid perpendicular to the $x$ axis are squares. Look down at the picture of the base and imagine that your solid is made up of many square cards, each of which has its bottom edge lying within your base (the semicircle). It may be hard to visualize what this pile of cards looks like exactly, but that's less important than the idea of an object made up of lots of vertical slices. If your hand isn't bothering you too much you can get some stiff paper and cut it into squares and prop them vertically on the floor.

*Once you have the picture, the rest is more or less easy. For every $x$ value from $-3$ to $3$, what is the edge size of its associated square? Easy: it's just $\sqrt{9-x^2}$. Again, look at Babak's picture.

*Now what will the area of each square be? You know that the side lying on the floor is of length $\sqrt{9-x^2}$, so the area will be $(\sqrt{9-x^2})^2 = 9-x^2$.

*All that's left is to integrate all of these areas, from $x=-3$ to $x=3$. We'll then have the volume equal to
$$
\int_{-3}^39-x^2\;dx
$$
and I'll leave the rest for you to do.
Hope this helps. Just remember, you can do this.
