I'm struggling with this question in my book. It wants me to find the surface integral
$$\iint_s F \cdot n \,dS.$$
It also says that I can observe something about the object to make it far less computationally intensive.
I am given $F = (x\ln(x^2+y^2),y\ln(x^2+y^2),0)$ and the the object.
The answer that I am supposed to receive is $4\pi R^2 h \ln(R)$.
Here is the work that I have done, but it never seems to work out to the right answer.
I observe that there is no force in the $\hat k$ direction and therefore the top and the bottom of the cylinder are irrelevant.
I then move on to set up the integral of the cylindrical surface without a top and bottom.
I determine the normal to be $(2x,2y,1)$ using the formula for the cylinder $x^2+y^2=R^2$.
Then dotting this and plugging it into the integral I get
$$\iint 2x^2\ln(x^2+y^2)+2y^2\ln(x^2+y^2)\, dA.$$
I then convert these into cylindrical coordinates to get
$$2\int_0^{2\pi}\int_0^r r^3cos^2(\theta)\ln(r^2)+r^3sin^2(\theta)\ln(r^2) \,drd\theta.$$
Pulling out constants I get
$$2r^2\ln(r^2)\int_0^{2\pi}\int_0^r r \,dA.$$
This evaluates to $4(\pi)^2r^2\ln(r^2)$, which is nowhere close to the right answer provided above. I am confused where I went wrong or what I am missing about the problem. I know I need $h$, but I don't know how this would factor into the equation at all.
It looks like you are integrating over the top or bottom disc. You should integrate over the cylinder mantle, which can be parametrized by $$ \Sigma(\theta, z) = (R\cos\theta, R\sin \theta, z) $$ with $\theta \in [0,2\pi]$ and $z \in [0,h]$. The normal is $n(\theta,z) = (R\cos \theta, R\sin \theta, 0)$. Also $$ F(\Sigma(\theta,z)) = (R\cos\theta \ln R^2, R\sin\theta\ln R^2,0), $$ so the integrand becomes $F(\Sigma(\theta,z))\cdot n(\theta,z) = R^2 \ln R^2 = 2R^2 \ln R$. Integrating this over the mantle is easy since it's constant w.r.t. to $\theta$ and $z$. The integral is $2\pi h (2R^2 \ln R)$, which is the answer.