So I have the following three questions (and no solutions to them sadly). I wanted to know if my results for a) and b) are correct and how I should proceed for c).
Given is the following ellipsoid : $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$$ and the following questions :
a) Calculate the volume of the ellipsoid
b) Calculate the flux of $V=(x-3xy^2,y^3-2xyz,xz^2+2z)$ through the surface of the ellispoid toward the exterior
c) Given is the cardiode $(x^2+y^2-2x)^2=4(x^2+y^2)$. Determine the "area which is enclosed by the cardioid" (german: Flächeninhalt).
a) As usual, we make the substitutions $\hat{x}=\frac{x}{a}$, $\hat{y}=\frac{y}{b}$ and $\hat{z}=\frac{z}{c}$, we get $dx=ad\hat{x}, dy=bd\hat{y}, dz = cd\hat{z}$, so now we have a sphere and can set up our triple integral in spherical coordinates $$abc \int_{0}^{1} \int_{0}^{2\pi}\int_{0}^{\pi}1*r^2sin(\phi)d\phi d\theta dr$$ which after integration yields the well known $abc \frac{4\pi}{3}$
b) The divergence of $V$ is simply 3. Thus we can use the divergence theorem : we have a triple integral (i.e. the volume of your ellipsoid) over the divergence of our vector field. In our case, the divergence doesn't depend on x, y or z, so we simply get our volume times our divergence, i.e. $abc \frac{4\pi}{3}*3=abc 4 \pi$
c) Now, here I really have some troubles, because I'm supposed to calculate the "area enclosed" using multivariable calculus, i.e. not just applying a well known formula. How am I supposed to proceed ? Should I use polar coordinates or what ?
Thanks for your help !
Edit 11/01/2019:
In fact, c) can be solved really easily. If we just use polar coordinates $x=r\cos(\theta), y=r\sin(\theta)$, we get $(r^2-2r\cos(\theta))^2=4r^2$. If we take the square root, the right side has to be positive because $r$ cannot be negative, so we get $2r$. For the left side, we have $\pm(r^2-2r\cos(\theta))=2r$ . So either $r(r-2\cos(\theta)-2)=0$ and either $r=0$ or $r=2(\cos(\theta)+1)$ or $r(-r+2\cos(\theta)-2)=0$ and $r=0$ or $r=2(\cos(\theta)-1)$ which is not possible because $r$ would be negative if not zero.
Thus $r=2(\cos(\theta)+1)$. Now we can set up our double integral:
$$\int_{0}^{2\pi}\int_{0}^{2(\cos(\theta)+1)}1 r dr d\theta=\int_{0}^{2\pi}2(\cos(\theta)+1)^2 d\theta =2 \int_{0}^{2\pi}(\cos^2(\theta)+2\cos(\theta)+1) d\theta=2(\pi+ 0 +2\pi)=6\pi$$
