The following question was recently asked in a lecture:
Using cylindrical polar coordinates find the area of the curved surface of a cone of height $h$ and radius $a$.
My attempt to do this was as follows:
By drawing a diagram, we can see that the equation of a cone in cylindrical Polars is $z= \frac{h}{a}R$ (where $0 \leq R \leq a$)
Now by integrating this, we get $\int_{0}^{2 \pi} \int_{0}^{a} \frac{h}{a}R \; dRd \theta$. Evaluating this integral gives the solution Area=$\pi a h$.
This solution is wrong. We have been given the answers and I should have got $\pi a \sqrt{a^{2}+h^{2}}$.
Why have I got this so wrong?
You're not integrating the right thing. It seems like you're trying to compute $\iint z \, dxdy$, except that you forgot the $R$ in $dxdy=R \, dRd\theta$. But this would not give the surface area, but the volume of the region below the cone and above the $xy$ plane. To get the surface area, you need to write down the expression for the surface area element and integrate that instead.