Question

The base of a solid is the region bounded by the parabolas y=x^2 and y=2-x^2. Find the volume of the solid if the cross-sections perpendicular to the x-axis are squares with one side lying along the base.

Answer 1

Well, this is not far from your last question:
http://openstudy.com/study#/updates/5254a429e4b037f05886a54d
It's actually even simpler, and same idea.

We have a base bounded by 2 given functions:
\[
y_1 = x^2 \\
y_2 = 2 - x^2
\]
We can find the difference in values of those two functions. Notice inside the base y2 > y1 ( though for area it doesn't matter as it is squared, so even negative difference value leads to the same result)
\[
\text{values_difference} = y_2 - y_1 = (2 - x^2) - x^2 = 2 - 2x^2
\]
We are told that once we slice the object into infinite number of slices along the X axis, we get squares with one side along the base.
That means that for a given X the matching slice's area would be the distance between both points of those 2 functions (which is the just the values difference) squared:
\[
\text{slice_area} = \text{values_difference}^2 = (2 - 2x^2)^2= \\
=4x^4 -8x^2 + 4
\]
Now, we wanna find the X values the base is in between. We can check where the functions intersect:
\[
\text{values_difference} = 0 \\
2 - 2x^2 = 0 \\
2 = 2x^2 \;// \div2 \\
x^2 = 1 \;//\sqrt{} \\
x = \pm 1
\]
Which means, our base is from x = -1 to x = 1.
Now again.. Sum up all slices between those X's by integration:
\[
\text{volume} = \int^1_{-1} \text{slice_area} \cdot dx = \\
= \int^1_{-1} (4x^4 - 8x^2 + 4) dx
= \huge[ \normalsize\frac{4x^5}{5} - \frac{8x^3}{3} + 4x \huge]^{\normalsize 1}_{\normalsize -1} \normalsize = \\
= \huge [ \small \frac{4 \cdot 1^5}{5} - \frac{8 \cdot 1^3}{3} + 4 \cdot 1 \huge ] \normalsize - \huge [ \small \frac{4 \cdot (-1)^5}{5} - \frac{8 \cdot (-1)^3}{3} + 4 \cdot (-1) \huge ] \normalsize =\\
= \huge [ \normalsize \frac{4}{5} - \frac{8}{3} + 4 \huge ] \normalsize - \huge [ \normalsize -\frac{4}{5} + \frac{8}{3} -4 \huge ] \normalsize = \\
= 2 \cdot \huge [ \normalsize \frac{12 - 40 + 60}{15} \huge ] \normalsize = \frac{64}{15}
\]
Hope that's right..