Question

Evaluate the ∫∫s y dS where S is the portion of the cylinder x2 + y2 = 3 that lies between z = 0 and z = 3.

Answer 1

Consider the surface $S$ , which is the portion of the cylinder $x^2+y^2=3$ lies between the planes $z=0$ and $z=3$ .

Let $x=\sqrt{3}\cos u$ , $y=\sqrt{3}\sin u$ and $z=v$ , where $0\leq u \leq2\pi, 0 \leq v \leq 3$

So, the parametric equation of surface can be represented as,

$$r(u,v)=⟨\sqrt{3}\cos u,\sqrt{3}\sin u,v ⟩$$

The normal to thee surface $S$ is,

$$r_u×r_v=\begin{Vmatrix} i & j & k \\ -\sqrt{3}\sin(u) & \sqrt{3}\cos(u) & 0\\ 0 & 0 & 1 \end{Vmatrix} =⟨\sqrt{3}\cos(u),\sqrt{3}\sin(u),0 ⟩$$

Now, find the surface integral as,

$\iint_S(y)dS=\iint_{uv}(\sqrt{3}\sin u)∥ r_u×r_v∥dudv$

$=\int_{0}^{2\pi}\int_{0}^{3}\sqrt{3}\sin u \sqrt{(\sqrt{3}\sin u)^2+(\sqrt{3}\cos u)^2}dvdu$

$=\int_{0}^{2\pi}\int_{0}^{3}\sqrt{3}\sin u (\sqrt{3})dvdu$

$=\int_{0}^{2\pi}\int_{0}^{3}3\sin u dvdu$

$=3[-\cos u ]_{0}^{2\pi}[ v]_{0}^{3}$

$=-3(1-1)(3-0)$

$=-3(0)(3)$

$=0$

Therefore, the surface integral is $\iint_S(y)dS=0$