The circle you see above is the equator of the sphere.
Its radius from the center is $r=3$ and $r^2=x^2+y^2$.
The segment running parallel to the $y$-axis is the diameter of a circular cross section of the solid sphere. Find the area $A(x)$ and volume $V(x)$ of the cross section.
My Answer:
The distance from the cross section to the end of the sphere is $r-y$ and $r=3$ so we have
$$3-y.$$
If $x^2 + y^2 = r^2$ then
\begin{align} y^2 = r^2 - x^2\\ y = \sqrt{r^2 - x^2}\\ y = Sqrt[9 - x^2]\\ \end{align} The area of a circle is determined by $\pi r^2$ so we can say the area function for a cross section that crosses the $x$-axis is
\begin{align} A(x) = \pi \sqrt{9 - x^2}^2\\ A(x) = \pi (9 - x^2)\\ \end{align} To get the volume of the cross section we can integrate from the corner $xy(3-y)$ to $x$.
$$V(x) = \int_{3-y}^{x}\pi (9 - t^2) dt$$
Why it is wrong:
This is not correct. How can you do this with a general radius of $r$? Also, consider you have solved for the $A(x)$ in terms of $x$. If you integrate from something to $x$, you'll get a function of $x$, not a formula just in $r$.
What is the correct way to do this given the reasons I am wrong?