I want to calculate the surface integral of $A=(4x,-2y^2,z^2)$ over the convex surface $S$ of the cylinder $x^2+y^2=4$ between $z=0$ and $z=3$. I get the unit normal vector $n$ as $(x/2,y/2,0)$. Now $A\cdot n=2x^2-y^3$.
I project the convex surface on the $xz$ plane and proceed to find $|n\cdot j|$ which comes out to be $y/2$. From the equation of the cylinder $y^2=4-x^2$ which leads to $y=\sqrt{4-x^2}$. Now
$$
\int_{S} A\cdot n dS= \iint (2x^2-y^3)/|n\cdot j| dxdz
= 2\iint((2x^2/y)- y^2) dxdz \\ = 2\iint ((2x^2/\sqrt{4-x^2})- 4+x^2) dxdz
$$
where the limits of integration are from $-2$ to $2$ in case of $x$ and $0$ to $3$ in case of $z$. When I evaluate this double integral in Wolframalpha, I get $24π-64$ as the answer. But my book evaluates it as $48π$. I am not able to understand where have I committed mistake. Please suggest.
Surface integral over the surface of a cylinder.
surface-integralsvector analysis
Best Answer
You only integrated over the half of the cylinder where $y>0$. You need to also include the integral for the half where $y < 0$. Then $y = -\sqrt{4-x^2}$, $|n\cdot j| = -y/2$, and the integral will come out to $24\pi + 64$. Add those together and you get $48\pi$.