I am a given a problem that reads "The base of $S$ is a region enclosed by $y = 2-x^2$ and the $x$-axis. Cross-sections perpendicular to the $y$-axis are quarter circles." The instructions are "Find the volume of the described solid S."
When I arrive at solving for $x$ from the above parabola, I get $x = \pm\sqrt{2-x^2}$. Now, my question is how do I find the area of my quarter circle? Since the area of a quarter circle is $\dfrac{1}{4}\pi r^2$, my trouble lies in understanding what '$r$' is in this case. Shouldn't the radius be $r = \sqrt{2-x^2}$? For some reason Chegg.com informs me $r = 2\sqrt{2-x^2}$. I'll link that page below, Thank you!
The radius, $r$ refers to the distance between the $y$-axis and the right half of the parabola. You can see this in the image given to you. So to find the radius, you will want to solve the equation for x. Since $x = \pm \sqrt{-y + 2}$. Hence $r = x = 2\sqrt{-y+2}$. The Area of the quarter circle then simplifies to $\frac{\pi}{4}4(-y+2)$. Finally, $\int_{0}^{2}\pi(-y+2)dy$ will give you the volume of the solid.