FORM GR1268 Question 22

Question:
What is the volume of the solid in \(xyz\)-space bounded by the surfaces \(y=x^2,y=2-x^2,z=0,\) and \(z=y+3\)?

(A) \(\displaystyle\frac{8}{3}\)     (B) \(\displaystyle\frac{16}{3}\)     (C) \(\displaystyle\frac{32}{3}\)     (D) \(\displaystyle\frac{104}{105}\)     (E) \(\displaystyle\frac{208}{105}\)

Answer:
(C)

Answer Key:
First, you want to picture this solid.

Take \(xy\)-plane parallel to the ground,
\(y\)-axis pointing to the right, and \(x\)-axis pointing toward you.
Draw \(y=x^2\) and \(y=2-x^2\).
The surrounded area look like eyelid or football.

Take \(z\)-axis pointing to the ceiling.
Extrude the eyelid along the \(z\)-axis.
Now we have a "column" whose cross-section is an eyelid.

Chop the column at \(z=0\) and \(z=y+3\).

We will use disc method for integration.

If we slice along the \(z\)-axis, we need the area of eyelid at each level of \(z\).
We'd rather avoid it, especially toward the top of the column,
where the cross-section is no longer an eyelid.

If we slice along the \(y\)-axis, each slice will be a rectangle.
At \(y=\tilde y\), the height of the rectangle is \(z=\tilde y+3\).

The width of the rectangle is a little tricky to calculate.
First note that \(y=x^2\) and \(y=2-x^2\) intersect at \(x=\pm1\).

For \(0\le \tilde y\le 1\), the width of the rectangle is the distance between the \(x\)-coordinates where \(y=x^2\) and \(y=\tilde y\) intersect.
That is, \(\tilde y=x^2\), so \(x=\pm\sqrt{\tilde y}\), so the width is \(\sqrt{\tilde y}-(-\sqrt{\tilde y})=2\sqrt{\tilde y}\).

For \(1\lt \tilde y\le 2\), the width of the rectangle is the distance between the \(x\)-coordinates where \(y=2-x^2\) and \(y=\tilde y\) intersect.
That is, \(\tilde y=2-x^2\), so \(x=\pm\sqrt{2-\tilde y}\), so the width is \(2\sqrt{2-\tilde y}\).

Then we integrate along the \(y\)-axis:
\[V=\int_0^12\sqrt{\tilde y}(\tilde y+3)d\tilde y+\int_1^2 2\sqrt{2-\tilde y}(\tilde y+3)d\tilde y,\]
which I'd rather not deal with.

If we instead slice the solid along the \(x\)-axis, each slice will be a trapezoid.

At \(x=\tilde x\), the width of the trapezoid is \((2-\tilde x^2)-x^2=2(1-\tilde x^2)\).
The \(y\)-coordinate of the left edge is \(y=\tilde x^2\), so the height is \(z=\tilde x^2+3\).
The \(y\)-coordinate of the right edge is \(y=2-\tilde x^2\), so the height is \(z=(2-\tilde x^2)+3=5-\tilde x^2\).
Thus, the are of the trapezoid is \(\frac{1}{2}[2(1-\tilde x^2)][(\tilde x^2+3)+(5-\tilde x^2)]=8(1-\tilde x^2)\).

Integrating along the \(x\)-axis yields
\[\begin{align}
V&=\int_{-1}^18(1-\tilde x^2)d\tilde x\\
&=8\int_{-1}^1(1-\tilde x^2)d\tilde x\\
&=8\left[x-\frac{\tilde x^3}{3}\right]_{-1}^1\\
&=8\left[\left(1-\frac{1}{3}\right)-\left(-1-\left(\frac{-1}{3}\right)\right)\right]\\
&=8\cdot\frac{4}{3}\\
&=\frac{32}{3}.
\end{align}\]