Q) The base of $S$ is the region enclosed by the parabola $y=1-x^2$ and the $x$-axis. Cross-sections perpendicular to the $y$-axis are squares.
I firstly set the area of cross-section as $1-y$, since the side of cross-section is $\sqrt{1-y}$.
Then, I just integrate $(1-y)$ from $0$ to $1$ with respect to $dy$, and I get $1/2$.
But I'm not sure whether this is correct.
Give me a help, please.
Thanks:)
Best Answer
Draw a picture. Note that the parabola is symmetric about the $y$-axis, so the base of $S$ is symmetric about the $y$-axis.
A cross-section at height $y$ is a square of side $2x$, where $x$ is the positive solution of $y=1-x^2$. So $x=\sqrt{1-y}$, and therefore the area $A(y)$ of cross-section is $4(1-y)$. Integrate from $y=0$ to $y=1$.