Science:Math Exam Resources/Courses/MATH101/April 2011/Question 02 (b)/Solution 1

If we follow the hint, we can draw a diagram similar to the one pictured on the right

. We have drawn the axis so that we see the $xy$ plane in its familiar 2D position.

The first thing we have to consider is the $x$ limits of where our picture begins and ends. To do this, we find the intersection points of our two parabolas. Therefore we want to find the $x$ such that
$\displaystyle {}x^{2}=2-x^{2}$
which occurs when $x=\pm {1}$. Notice from our diagram that $x=-1$ is the left endpoint and $x=1$ is the right endpoint. As we are told in the question, the cross section in the $z$-direction is a square which we can think of as a rectangular prism with a very small depth, ${\textrm {d}}x$. Therefore, the volume of any one square, ${\textrm {d}}V$, is given by
$\displaystyle {}{\textrm {d}}V=bh{\textrm {d}}x.$
Since the cross section is indeed a square we have that $b=h$ and this equals the height between the two parabolas. Therefore,
$\displaystyle {}b=h=(2-x^{2})-x^{2}=2-2x^{2}$
and thus for any $x$ we have that the volume of the rectangle is,
$\displaystyle {}{\textrm {d}}V=h^{2}{\textrm {d}}x=(2-2x^{2})^{2}{\textrm {d}}x=4(1-x^{2})^{2}{\textrm {d}}x.$
To get the total volume we have to add up all the little square volumes that we would sweep out from the start of our figure at $x=-1$ to the end at $x=1$. However, there are an infinite number of $x$ values in this range and so what would normally be a sum, we write as an integral. Therefore,
${\begin{aligned}V&=\int {\textrm {d}}V=\int _{-1}^{1}4(1-x^{2})^{2}{\textrm {d}}x\\&=4\int _{-1}^{1}(x^{4}-2x^{2}+1){\textrm {d}}x\\&={\frac {64}{15}}.\end{aligned}}$
Therefore, the volume of the solid is $V={\frac {64}{15}}$.