Compute Pool Volume: Angeezzzz's Question at Yahoo Answers

[MarkFL]

Here is the question:

Volume integration help?


As viewed from above, a swimming pool has the shape of the ellipse given by
x^2/3600+y^2/2500=1

The cross sections perpendicular to the ground and parallel to the y-axis are squares. Find the total volume of the pool. (Assume the units of length and area are feet and square feet respectively. Do not put units in your answer.)

V= ? ft^3

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Angeezzzz,

We first should write the ellipse in standard form:

\(\displaystyle \frac{x^2}{60^2}+\frac{y^2}{50^2}=1\)

Thus, we see the length of the semi-major axis is 60. We may restrict ourselft to the first quadrant, and then by symmetry quadruple the result to get the total volume. The volume of an arbitrary rectangular slice is:

\(\displaystyle dV=bh\,dx\)

where:

\(\displaystyle b=y=\frac{\sqrt{3000^2-50^2x^2}}{60}\)

\(\displaystyle h=2y=\frac{\sqrt{3000^2-50^2x^2}}{30}\)

Hence, we may state:

\(\displaystyle dV=\frac{3000^2-50^2x^2}{1800}\,dx=-\frac{25}{18}\left(x^2-3600 \right)\,dx\)

And so the total volume is given by:

\(\displaystyle V=-\frac{50}{9}\int_0^{60}x^2-3600\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle V=-\frac{50}{9}\left[\frac{x^3}{3}-3600x \right]_0^{60}=-\frac{50\cdot60^3}{9}\left(\frac{1}{3}-1 \right)=-\frac{50\cdot60^3}{9}\left(-\frac{2}{3} \right)=800000\)