Calc 3, area of an ellipsoid slice

[Allenman]

This isn't homework or anything, I just want to understand the question better.

Homework Statement

The Attempt at a Solution

I'm honestly not sure where to go with this. Is this an integral problem? As I understand it I'm finding the area of a slice, not a volume of the whole ellipsoid.

so z = c which = 3...
the problem then becomes:
x^{2} + \frac{y^{2}}{4} = 0

Does the last description mean that the answer is A= (\pi)(1)(2) ?
It doesn't seem like it should be that simple. lol

 

[Mark44]

This isn't homework or anything, I just want to understand the question better.

Homework Statement

The Attempt at a Solution

I'm honestly not sure where to go with this. Is this an integral problem? As I understand it I'm finding the area of a slice, not a volume of the whole ellipsoid.

so z = c which = 3...
the problem then becomes:
x^{2} + \frac{y^{2}}{4} = 0

Does the last description mean that the answer is A= (\pi)(1)(2) ?
It doesn't seem like it should be that simple. lol

No, because x² + y²/4 = 0 is not the equation of an ellipse.

Notice that there aren't many points that satisfy this equation.

Have you drawn a sketch of the ellipsoid? A sketch would help illuminate what's going on here.

 

[SammyS]

No, it's not an integral problem.

The only real solution for \displaystyle x^2+\frac{y^2}{4}=0 is x = ? = y .

Set z = c , then solve for \displaystyle x^2+\frac{y^2}{4}\,, then divide both sides by the right hand side.

 

[Mark44]

Where does it say that the plane is z = 3?

 

[Allenman]

But isn't 9 supposed to be c^{2}?

so really I have:
x^{2} + \frac{y^{2}}{4} + \frac{c^{2}}{c^{2}} = 1
which just leads me back to the previous equation where the only x and y values would be zero.
or am I just making a false assumption?

am I supposed to be doing this?
\frac{c^{2}}{9} = 1 - x^{2} - \frac{y^{2}}{4}

c = \sqrt{\frac{1 - x^{2} - \frac{y^{2}}{4}}{9}}

I don't know where i would go from there.The equation my instructor gave us for a ellipsoid is:
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = d
That's where I got c=3

 

[Mark44]

I'm not sure, but I don't think that c as used in this problem is related to the c in your ellipsoid equation. I believe they are using in the equation z = c to mean a plane that is parallel to the x-y plane, that's all.

 

[Allenman]

I think you're right Mark. I feel kind of silly for making that assumption. lol.

The book says the answer is:

\frac{2\pi(9 - c^{2})}{9}

I honestly don't see how it got that answer.

Thanks for all the help so far!

 

[Mark44]

You're looking for the area of the elliptical cross-section that is cut by the plane z = c.

Follow the directions that SammyS gave back in post #3, substituting c for z in the equation to get

x² + y²/4 = <something with c in it>

Now divide both sides by what you have on the right side to get this equation into the standard form for an ellipse. You will then be able to pick out a and b, which you can use in your area formula for an ellipse.