# Thread: Finding maximum volume of an ellipse (Optimisation under differentiation)

1. ## Finding maximum volume of an ellipse (Optimisation under differentiation)

OK
So, I have half an ellipse with the equation y = sqrt(1296-x^2). Here is the graph:
It is actually a curved 3d structure with the height 36m, so I found out the above equation by using the equation of the ellipse.
The question is:
To find the maximum volume of this structure (or curve)?

I can find the area by using integration but that wouldnt be the same as volume?
Any ideas?

PLS PLS HELP

PS. THE BUILDING LOOKS LIKE THIS:
Is it fair to say that it is an ellipse (half)?

[IMG]file:///C:/Users/Ameya/AppData/Local/Temp/moz-screenshot-7.png[/IMG]

THANK YOU SO MUCH!!!!

2. You'll need to use a triple integral (at least, that's the easiest way imo). I would do it in the order of dydxdz, but that's up to you. Your limits on y would be 0 to sqrt(1296-x^2), your limits on x would be found be using 0 = sqrt(1296 - x^2) and solving for x, and your limits on z would be -36 to 36. I believe this would be the correct answer, someone correct me if I'm wrong.

3. It would make more sense to call that a semi-circle.

4. Actually, I just thought of this - isn't that building more of 1/4 of an ellipse? It's cut in half by the ground, but it also seems to be cut in half again just by design on the front.

5. Well, the building is not the design! And, again, it is not an ellipse, it is a circle: If $y= \sqrt{1296- x^2}$ then $y^2= 1296- x^2$ so $x^2+ y^2= 1296= 36^2$. That is a circle with center at (0, 0) and radius 36. But we are talking about volume and for that we need three dimensions. What happens behind the semi-circle that we see? Is that a volume of rotation (hemisphere) or does it go straight back- a semi-circular cylinder?

The volume of a hemisphere is $\frac{2}{3}\pi r^3$, about 73287 for r= 36. The volume of a semi-circular cylinder is the area of the semi-circle, $\frac{1}{2}\pi r^2$, about 2035 for r= 36, multiplied by the depth of the cylinder that we do not know.

Actually, the shape in the photo looks more like part of a hyperbola to me.