MHB Compute Pool Volume: Angeezzzz's Question at Yahoo Answers

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The discussion revolves around calculating the volume of a swimming pool shaped like an ellipse, defined by the equation x^2/3600 + y^2/2500 = 1. The cross-sections of the pool, taken perpendicular to the ground and parallel to the y-axis, are squares. By analyzing the geometry, the volume is derived using integration, focusing on the first quadrant and then multiplying the result by four for symmetry. The final calculated volume of the pool is 800,000 cubic feet. This solution illustrates the application of calculus in determining volumes of irregular shapes.
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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.
 
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Hello Angeezzzz,

We first should write the ellipse in standard form:

$$\frac{x^2}{60^2}+\frac{y^2}{50^2}=1$$

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:

$$dV=bh\,dx$$

where:

$$b=y=\frac{\sqrt{3000^2-50^2x^2}}{60}$$

$$h=2y=\frac{\sqrt{3000^2-50^2x^2}}{30}$$

Hence, we may state:

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

And so the total volume is given by:

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

Applying the FTOC, we obtain:

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