calculusintegrationmultivariable-calculus
I have to integrate over the curved surface of a inverted cone, since the surface is of a cone I think all three $s, \phi$ and $z$ are varying under the integral sign.
Now this causes the problem, I don't know surface element when all three are varying nor I can find a solution on internet, I thought I will take $\displaystyle|d\vec a| = s\ ds\ d\phi\ dz$ but that is volume element.
What should I do ? Drop $s$ from $\displaystyle |d\vec a| = s\ ds\ d\phi \ dz$ ?
I've come across the picture you're looking for in physics textbooks before (say, in classical mechanics). A bit of googling and I found this one for you!
Alternatively, we can use the first fundamental form to determine the surface area element. Recall that this is the metric tensor, whose components are obtained by taking the inner product of two tangent vectors on your space, i.e. $g_{i j}= X_i \cdot X_j$ for tangent vectors $X_i, X_j$. We make the following identification for the components of the metric tensor, $$ (g_{i j}) = \left(\begin{array}{cc} E & F \\ F & G \end{array} \right), $$ so that $E = <X_u, X_u>, F=<X_u,X_v>,$ and $G=<X_v,X_v>.$
We can then make use of Lagrange's Identity, which tells us that the squared area of a parallelogram in space is equal to the sum of the squares of its projections onto the Cartesian plane: $$|X_u \times X_v|^2 = |X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2.$$
Here's a picture in the case of the sphere:
This means that our area element is given by $$ dA = | X_u \times X_v | du dv = \sqrt{|X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2} du dv = \sqrt{EG - F^2} du dv. $$
So let's finish your sphere example. We'll find our tangent vectors via the usual parametrization which you gave, namely, $X(\phi,\theta) = (r \cos(\phi)\sin(\theta),r \sin(\phi)\sin(\theta),r \cos(\theta)),$ so that our tangent vectors are simply $$ X_{\phi} = (-r\sin(\phi)\sin(\theta),r\cos(\phi)\sin(\theta),0), \\ X_{\theta} = (r\cos(\phi)\cos(\theta),r\sin(\phi)\cos(\theta),-r\sin(\theta)) $$ Computing the elements of the first fundamental form, we find that $$ E = r^2 \sin^2(\theta), \hspace{3mm} F=0, \hspace{3mm} G= r^2. $$ Thus, we have $$ dA = \sqrt{r^4 \sin^2(\theta)}d\theta d\phi = r^2\sin(\theta) d\theta d\phi $$
Since you are given $$z=c\sqrt {x^2+y^2}= cr$$ and you know that $$\tan \phi = \frac {r}{z}$$
We get $$\tan \phi =1/c$$
Thus the limits for $\phi$ are $$0\le \phi \le \tan ^{-1} (1/c)$$
Best Answer
A surface integral needs two parameters although the surface element vector needs three components. We can write those components in cartesian coordinates and the surface parameters can be adopted from the description in cylindrical coordinates.
For an inverted cone: $z^2=K(x^2+y^2)$ (with $K$ some constant), that in cylindrical is, $z=\pm K s$, dropping the minus sign as it's stated that it's an inverted cone, $z=K s$. We can parametrize the surface with the help of the cylindrical coordinates $s$ and $\phi$ as $z$ is determined by the equation for the cone: In cartesian coordinates, the parametric equations for this cone are:
$$C:r=(x,y,z)=(s\cos\phi,s\sin\phi,Ks)$$
Now, to find the pointing outwards, normal vector, we need two vectors tangent to the surface, then the former is the cross product of the laters:
$$\begin{cases} \dfrac{\partial r}{\partial\phi}=(-s\sin\phi,s\cos\phi,0)\\ \dfrac{\partial r}{\partial s}=(\cos\phi,\sin\phi,K) \end{cases}$$
$$\dfrac{\partial r}{\partial\phi}\times\dfrac{\partial r}{\partial s}=(Ks\cos\phi,Ks\sin\phi,-s\sin^2\phi-s\cos^2\phi)=$$
$$=s(K\cos\phi,K\sin\phi,-1)$$
And
$$\left|\dfrac{\partial r}{\partial\phi}\times\dfrac{\partial r}{\partial s}\right|=s\sqrt{K^2+1}$$
The integral is in cylindrical coordinates,
$$\iint_S f(s,\phi)s\sqrt{K^2+1}\,dsd\phi$$