MATH101 April 2011
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Question 02 (b)

Consider a solid whose base in the xyplane is the finite region bounded by the curves $y=x^{2}$ and $y=2x^{2}$. The cross sections of the solid perpendicular to the xaxis with one side in the xyplane are squares. Find the volume of this solid.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Draw a picture. What are the limiting values on the xaxis?

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Solution

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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
$\displaystyle {}x^{2}=2x^{2}$
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,
$\displaystyle {}b=h=(2x^{2})x^{2}=22x^{2}$
and thus for any $x$ we have that the volume of the rectangle is,
$\displaystyle {}{\textrm {d}}V=h^{2}{\textrm {d}}x=(22x^{2})^{2}{\textrm {d}}x=4(1x^{2})^{2}{\textrm {d}}x.$
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,
${\begin{aligned}V&=\int {\textrm {d}}V=\int _{1}^{1}4(1x^{2})^{2}{\textrm {d}}x\\&=4\int _{1}^{1}(x^{4}2x^{2}+1){\textrm {d}}x\\&={\frac {64}{15}}.\end{aligned}}$
Therefore, the volume of the solid is $V={\frac {64}{15}}$.

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