# Surface integral with cylinder-form surface.

Find the surface integral

$$\iint_S F(x,y,z)$$ if $F(x,y,z)=x+y+z$ and $S: x^2 + y^2 = 1 , 0 \leq z \leq 2$.

Obviously, this is the first type of surface integral since the function $F$ is scalar field, also, surface given is a piece of cylinder whose height is two, and this actually isn't problem at all, the problem occurs once i try to implement this.

By definition first type of surface integral is given as $$\iint_S F(x,y,z)=\iint_DF(x,y)\sqrt{1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2}dxdy$$

We don't actually have to eliminate $z$ variable every time, it actually differs from one case to another, anyway, i am having trouble with expressing $z$ in terms of two other variables since $z$ is not dependent on $x$ and $y$, i cannot express $x$ in terms of $y$ since $y=\sqrt{1-x^2}$ and $y=-\sqrt{1-x^2}$ so it gives me two distinct functions, same works for $y$ variable.

When i saw i could not do anything i mentioned above i tried introducing cylindric coordinates but then i got $$S: r=1, 0\leq z \leq 2$$ i could not go further. Anyone knows how to solve this?

• Use $x=r\cos\theta,y=r\sin\theta,z=z$ and express the surface integral in terms of cylindrical coordinates – PiGamma Feb 3 '18 at 13:39
• @PiGamma I actually tried it, as i mentioned in my post, but i could not get the rhs expression of the surface integral definition which is necessary in order to solve it. – cdummie Feb 3 '18 at 13:46

Hint: Let a surface $S$ has vector equation $$\vec{r}(u,v)=x(u,v)\vec{i}+y(u,v)\vec{j}+z(u,v)\vec{k}$$$(u,v)\in D$ where $D$ is the parameter domain.Then $$\iint_S f(x,y,z)dS=\iint_Df(\vec{r}(u,v))|\vec{r_u} \times\vec{r_v}|dA$$
• ok i have this vector equation: $\vec{r}(u,v)=cos\theta \vec{i}+sin\theta \vec{j}+z\vec{k}$ which still leaves me without a possibility to somehow find a relation between $z$ and other two variables which is necessary if i know that $F=x+y+z$ – cdummie Feb 3 '18 at 16:05
• Substitute the vector equation in $F$ – PiGamma Feb 4 '18 at 3:12
• Thanks, i did that and now i have $F=sin\theta + cos\theta + z$ which actually is good, because now i have $F$ in terms of two variables, which is what i need for rhs of the equation, but now i don't know what to do about $\sqrt{1+(\frac{\partial f}{\partial z})^2+(\frac{\partial f}{\partial \theta})^2}$, and what about Jacobian, is it necessary here? Back then, when i learned about double integrals i had to calculate jacobian for every substitution i introduce. – cdummie Feb 4 '18 at 7:48
• Yes, of course, you should calculate Jacobian which is $|\vec{r_\theta} \times\vec{r_z}|$ in this case – PiGamma Feb 4 '18 at 12:50