Volume with Cross Sections Perpendicular to Y-Axis

Question

Find the volume of the solid whose base is the region enclosed by y=x^2 and y=1, and the cross sections perpendicular to the y-axis are squares.

Anyone know why the answer is 2?

Answer 1

The volume you are integrating over should look like a pyramid with curved sides, specifically, $y=z^2$. $$ 2\int^{1}_{-1}\int^{1}_{x^2}\int^{\sqrt{y}}_{0}dz dy dx $$ Since the volume is symmetric I integrated over positive z and multiplied the volume by 2. I recommend you google perpendicular cross section as that will help visually solidify what has been done.

Answer 2

Turn your view of the x-y axis to the y-x axis.

You have the area bounded by $x=\sqrt{y}$, $x=-\sqrt{y}$, and $y=1$.

So we have our bounds of integration are $(0,1)$.

The side length of the square for each $y$ are $2\sqrt{y}$.

Our volume is the sum of several squares with infinitesimal width (dy).

We have $\displaystyle \int_0^1 \left(2\sqrt{y}\right)^2\,dy=\boxed{2}$