An ellipsoidal meteor is careening down from outer space. The comet's exterior takes the shape of $\frac{(x-30)^2}{36}+\frac{(y-70)^2}{9}+\frac{(z-40)^2}{25}=1$. When the meteor hits the Martian troposphere, the temperature at a given (x,y,z)(units in feet) on a comet is $h(x,y,z)=\sin(x)+\cos(y)+5yz$, in degrees (F). Compute the average temperature of the exterior of the comet at this moment. (Use surface area approximation)
My attempt: Integrate the temperature around "rings" perpendicular to the $z$-axis, so we hold $z$ constant, and take a parametric integral of the cross section. We then integrate this expression all the way up from $z=35$ to $z=45$. Then we divide by the surface area of the ellipsoid.
So I begin with redefining the comet so that $\frac{x^2}{36}+\frac{y^2}{9}+\frac{z^2}{25}=1$ and $H(x,y,z)=\sin(x+30)+\cos(y+70)+(y+70)(z+40)$.
Then my rings will start at $z=-5$ up to $z=5$. Then I will have $\frac{x^2}{36}+\frac{y^2}{9}=1-\frac{z^2}{25}$, so the normal 'radius' of $x$ is 6, and 3 for $y$, and the scalar (of the elliptical cross section with respect to the ellipse at the diameter) for each $z$ will be $\sqrt{1-\frac{z^2}{25}}$.
So we have $x(t)=6\sqrt{1-\frac{z^2}{25}}\cos(t)$, and $y(t)=3\sqrt{1-\frac{z^2}{25}}\sin(t)$.
Therefore, I have the temperature of the ring is $\displaystyle \scriptsize{\int_{-5}^{5} \int_0^{2\pi}\left[\sin\left(6\sqrt{1-\frac{z^2}{25}}\cos(t)+30\right)+\cos\left(3\sqrt{1-\frac{z^2}{25}}\sin(t)+70\right)+\left(3\sqrt{1-\frac{z^2}{25}}\sin(t)+70\right)\left(z+40\right)\right]\,dt\,dz}$.
But I have to divide by the surface area of the ellipse.
The approximation formula gives $4π\large{\sqrt[1.6075]{\frac{(ab)^{1.6075}+(bc)^{1.6075}+(ac)^{1.6075}}{3}}}$, so if I plug in $a=6,b=3,c=5$, this is equal to $271.3253$ ft$^2$. So, my final integral is
$\displaystyle \tiny{T=\ \frac{1}{271.3253}\int_{-5}^{5} \int_0^{2\pi}\left[\sin\left(6\sqrt{1-\frac{z^2}{25}}\cos(t)+30\right)+\cos\left(3\sqrt{1-\frac{z^2}{25}}\sin(t)+70\right)+5\left(3\sqrt{1-\frac{z^2}{25}}\sin(t)+70\right)\left(z+40\right)\right]\,dt\,dz}$.
Finally, we have that $\boxed{T=3218.328 Fº}$
This makes sense, as I just read on google that the average temperature of a meteor is $3000 Fº$.
I wanted to make sure this is correct. This is for pre Calc, and the worksheet is worth a lot of points, and the problems are kind of hard.
