Question

show that surface a integral =48pi where a=4xi-2y^2j+z^2k and s is curved surface of cylinder x^2+y^2=4 bounded by plane z=0 to 3

Answer 1

Let $x=2\cos\varphi$ and $y=2\sin\varphi$, so we have $s=\{(\varphi,z)|0\le\varphi\le 2\pi, 0\le z\le 3\}$

$(x'_{\varphi},y'_{\varphi},z'_{\varphi})=(-2sin\varphi,2\cos\varphi,0)$ and $(x'_z,y'_z,z'_z)=(0,0,1)$.

We have that normal to the cylinder is $\vec{N}=(x'_{\varphi},y'_{\varphi},z'_{\varphi})\times(x'_z,y'_z,z'_z)=$

$=(-2sin\varphi,2\cos\varphi,0)\times(0,0,1)=(2\cos\varphi,2\sin\varphi,0)$

So $\iint\limits_s \vec{a}d\vec{s}=\iint\limits_s (\vec{a},\vec{N})d\varphi dz=\int\limits_0^3\int\limits_0^{2\pi}(\vec{a},\vec{N})d\varphi dz$

We have $\vec{a}=(4x,-2y^2,z^2)=(8\cos\varphi,-8\sin^2\varphi,z^2)$, so $(\vec{a},\vec{N})=((8\cos\varphi,-8\sin^2\varphi,z^2),(2\cos\varphi,2\sin\varphi,0))=$

$=16\cos^2\varphi-16\sin^3\varphi$

Since $\cos^2\varphi=\frac{1+\cos 2\varphi}{2}$ and $\sin^3\varphi=\frac{3\sin\varphi-\sin 3\varphi}{4}$, we have $(\vec{a},\vec{N})=8+8\cos 2\varphi-12\sin\varphi+4\sin 3\varphi$

Then $\iint\limits_s \vec{a}d\vec{s}=\int\limits_0^3\int\limits_0^{2\pi}(\vec{a},\vec{N})d\varphi dz=$

$\int\limits_0^3\int\limits_0^{2\pi}(8+8\cos 2\varphi-12\sin\varphi+4\sin 3\varphi)d\varphi dz=$

$=\int\limits_0^316\pi dz=48\pi$