Area of a given interval and volume of an ellipsoid

[skate_nerd]

I am given a pretty basic ellipsoid:
$$\frac{x^2}{16}+\frac{y^2}{9}+\frac{z^2}{1}=1$$
First, for each number t in the interval \(-1\leq{t}\leq{1}\) I need to find the area A(t) of the plane cross-section made by \(z=t\). This I know should be a function of \(t\).
After that I have to find the volume of this ellipsoid.

So far I have a sketch of this ellipsoid, which was pretty easy because it's just each denominator's square root is the length of the ellipsoid on that axis, respectively. However in my class so far we have only covered integrals and differentiation regarding vectors in 3-space. However to find an area, you need an integral, correct? I guess the bounds of the integral would probably be -1 to 1, but how would I go about finding a function to integrate? I'm assuming you would have to probably put this \(z=t\) in the ellipsoid's equation, but from there would I need to do something like solve for y? I guess I just need a little help getting started.

 

[MarkFL]

To get you started, if the ellipse is given by:

$\displaystyle \frac{x^2}{4^2(1-t^2)}+\frac{y^2}{3^2(1-t^2)}=1$

then its area is:

$\displaystyle A_t=\pi(4^2(1-t^2))(3^2(1-t^2))=144\pi(1-t^2)^2$

and so the volume of the ellipsoid would be:

$\displaystyle V=\int_{-1}^1 A_t\,dt$

 

[skate_nerd]

For the equation of that ellipse, where did the \(1-t^2\)'s on the denominators come from? Did you factor out the \(t^2\) from the polynomial or something?

 

[MarkFL]

First, I let $z=t$ to get:

$\displaystyle \frac{x^2}{4^2}+\frac{y^2}{3^2}+t^2=1$

Subtract through by $t^2$:

$\displaystyle \frac{x^2}{4^2}+\frac{y^2}{3^2}=1-t^2$

Divide through by $1-t^2$:

$\displaystyle \frac{x^2}{4^2(1-t^2)}+\frac{y^2}{3^2(1-t^2)}=1$

Then I used the formula for the are of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

which is:

$A=\pi ab$.

Now, I made a mistake above...I should have written:

$A_t=\pi\sqrt{4^2(1-t^2)}\sqrt{3^2(1-t^2)}=12\pi(1-t^2)$

I did not properly apply the formula. (Tmi)