I'm asked to calculate the surface area of $$x^2 + y^2 = 4, \quad 0 \leq z \leq 3$$ using the surface integral.
This is my attempt:
For symmetry reasons I calculate the surface area when $y>0$ and then multiply by 2.
For $y>0$ This surface can be parametrized by $$(x,z) : (x, \sqrt{4-x^2}, z)$$
I set $$r(s,t) = (2\cos{s}, 2\sin{s}, t) \quad 0 \leq s \leq 2\pi ,\quad 0 \leq t \leq 3$$ I find the normal to the surface expressed in $(s,t)$ to be $$(2\cos{s}, 2\sin{s},0)$$ I find the Jacobian of the transformation $(x,z) --> (s,t)$ to be $-2\sin{s}$. Using the definition of surface integral I get: $$\int\int\sqrt{4\sin^2{s} + 4\cos^2{s}} \cdot -2\sin{s} \quad dsdt$$ $$-4\int\int \sin{s} \quad dsdt$$ $$-4\int_{0}^{\pi}\sin{s}\quad ds \cdot \int_0^3 1\quad dt= -24$$ Now multiplying by 2 (to get both sides) I get -48 (I understand the answer can't be negative). And the answer is supposed to be $12\pi$. What am I doing wrong?
This problem is easier done in cylindrical coordinates. The surface has parametrization $$(2\cos(\theta),2\sin(\theta),z)$$ where $0\leq \theta\leq 2\pi$ and $0\leq z \leq 3$. The jacobian of the transformation is then $2$. The surface integral becomes $$\int_0^{2\pi}\int_{0}^32dz\,d\theta = 12\pi$$ Your jacobian seems to be the one for spherical coordinates. Double check your jacobian calculation
Edit. For these types of problems, I like to consider the full cylindrical coordinated map $(r,\theta,z)\mapsto (r\cos(\theta),r\sin(\theta),z)$. The jacobian for this matrix is standard to compute and gives $r$. If you remember $r^2 = x^2+y^2 = 4$, then you get the jacobian I computed above. Try to use these standard transformations, e.g. cylindrical, spherical, linear, etc., since it is easy to check if you have your computations correct