What is the surface integral of hemisphere?

Question

Calculate the integral $∫ ∫_{Y}z dS$ if Y is the part of the conic surface $z = \sqrt{x^2 + y^2}$ where z is between 0 and 1.

So this is what I did: $z = \sqrt{4-x^2-y^2}$ and then I used spehere's coordinates and got z to $z=2\sqrt{1-\sin^2(\theta)}$ and my area elemnent or whatever it is called is given by $r^2\sin(\theta)$ and then I multiplied this by z and parted the integrals. I for some reasons think that I don't need to take r as the variable(I saw a similar quation where someone just took r as a constant, don't understand why and if I can do that here). This means my variables are theta and phi and the bounds are $0\leq \theta \leq \frac{\pi}2$ and $0\leq \phi \leq 2\pi$. First of all I don't know if I have done right so far so correct me if I'm wrong but otherwise how in the hell am I supposed integrate $\int 2\sqrt{1-\sin^2(\theta)} = 2 \int \sqrt{\frac12 + \cos(2\theta)}$

EDIT: I accidentally posted the wrong question, this was the question my "solution" was for: Evaluate the surface integral $\int\int_{Y}zdS$ when Y is the upper half of the sphere $x^2+y^2+z^2=4$, i.e. the part where z ≥ 0.

Answer 1

First see that the surface is $z=\sqrt{x^{2}+y^{2}}=f(x,y)$ .

Then the surface area element $dS$ when projected to the $x,y$ plane would be $\sqrt{1+(\frac{\partial f}{\partial x})^{2}+(\frac{\partial f}{\partial y})^{2}}\,dxdy$

Which is equal to $\sqrt{1+(\frac{x}{\sqrt{x^{2}+y^{2}}})^{2}+(\frac{y}{\sqrt{x^{2}+y^{2}}})^{2}}\,dxdy=\sqrt{2}\,dxdy$

So the integral is $\int_{D}\sqrt{2}\sqrt{x^{2}+y^{2}}\cdot\,dx\,dy$ where $D$ is the disk $x^{2}+y^{2}\leq 1$

Now use polar coordinates to get $\int_{0}^{2\pi}\int_{0}^{1}\sqrt{2}\cdot r\cdot r\,dr\,d\theta=\frac{2\sqrt{2}\pi}{3}$

EDIT :- Well as per the question that you want to ask , see that now $z=\sqrt{4-x^{2}-y^{2}}=f(x,y)$ .

So again, the surface area element can be calculated as ,

$\sqrt{1+\frac{x^{2}+y^{2}}{4-x^{2}-y^{2}}}=\frac{2}{\sqrt{4-x^{2}-y^{2}}}$

So the surface integral over the $x,y$ plane now becomes,

$$\iint_{D}\frac{2}{\sqrt{4-x^{2}-y^{2}}}\cdot\sqrt{4-x^{2}-y^{2}}\,dx\,dy$$ where $D$ is the disk $x^{2}+y^{2}\leq 4$ .

$$=\iint_{D}2\cdot\,dxdy=2\cdot\text{Area of disk}=2\cdot(\pi 2^{2})=8\pi$$

Answer 2

I would recommend solving this problem in terms of cylindrical coordinates. First, recall that the scalar surface integral is $$ \iint_Sf\,dS=\iint_{\vec R}f(\vec R(u,v))|\vec N|\,du\,dv$$ where $\vec R(u,v)$ is the parametrization of the surface in terms of some variables $u,v$ and $\vec N$ is the normal vector. To solve this problem, let's first parametrize our cone in cylindrical coordinates. Recall that in terms of cylindrical coordinates $$ x=r\cos(\theta), \qquad y=r\sin(\theta), \qquad z=z $$ so we see that the cone $z=\sqrt{x^2+y^2}$ becomes $z=r$. Thus, we see that the cone can be parametrized as $\vec R(r,\theta)=(r\cos(\theta),r\sin(\theta),r)$ with $0\leq r\leq 1$ and $0\leq \theta\leq 2\pi$. Now, we must calculate the normal vector, so do this we calculate the tangent vectors $$e_r=(\cos(\theta),\sin(\theta),1)$$ $$ e_\theta=(-r\sin(\theta),r\cos(\theta),0)$$ and therefore the normal vector becomes $$\vec N= e_r\times e_\theta=(-r\cos(\theta),r\sin(\theta),r)$$ with the resulting magnitude $|\vec N|=\sqrt{2}r$. Thus, we have all we need to solve this integral. We get $$\iint_{\vec R}f(\vec R(r,\theta))|\vec N|\,dr\,d\theta=\int_0^{2\pi}\int_0^1r(\sqrt{2}r)\,dr\,d\theta=2\sqrt{2}\pi\left(\frac{r^3}{3}\right]_0^1=\dfrac{2\sqrt{2}\pi}{3}.$$ Hope this helps! Also I am kinda notorious for arithmetic errors so please let me know if I make an error.