Question

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Answer 1

Q:Find \[\iint_\limits S (\vec \nabla \times \vec F).\hat n \space ds\]
where
\[\vec F=(x-z)\hat i+(x^3+yz)\hat j+3xy^2 \hat k\]
and S is the surface of the cone
\[z=a-\sqrt{x^2+y^2}\]
above the xy plane
Now
\[\vec \nabla \times \vec F=\left|\begin{matrix}\hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\x-z & x^3+yz & 3xy^2\end{matrix}\right|\]\[\vec \nabla \times \vec F=\hat i(6xy-y)-\hat j(3y^2+1)+3x^2 \hat k=y(6x-1)\hat i-(3y^2+1)\hat j+3x^2 \hat k\]
We can re-write our equation for cone as
\[\sqrt{x^2+y^2}+z=a\]
Which is of the form
\[\phi(x,y,z)=c\]
Equation of a level surface, where phi is a scalar function of x,y,z and c is a constant
In this case, a vector normal to the surface is given by
\[\vec n=\vec \nabla \phi=(\hat i \frac{\partial \phi}{\partial x}+\hat j \frac{\partial \phi}{\partial y}+\hat k \frac{\partial \phi}{\partial z})\]\[\vec n=\frac{1}{2\sqrt{x^2+y^2}}.2x \hat i+\frac{1}{2\sqrt{x^2+y^2}}.2y \hat j+\hat k=\frac{x \hat i}{\sqrt{x^2+y^2}}+\frac{y \hat j}{\sqrt{x^2+y^2}}+\hat k\]
\[|\vec n|=\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1^2}=\sqrt{\frac{x^2+y^2}{x^2+y^2}+1}=\sqrt{1+1}=\sqrt{2}\]
\[\hat n=\frac{\vec n}{|\vec n|}=\frac{1}{\sqrt{2}}(\frac{x \hat i}{\sqrt{x^2+y^2}}+\frac{y \hat j}{\sqrt{x^2+y^2}}+\hat k)\]
When
\[z=0\]
We have
\[x^2+y^2=a^2\]
This is the projection of the cone in xy-plane (circle of radius a in the xy-plane),

Thus we convert our surface integral...
\[\iint_\limits S (\vec \nabla \times \vec F). \hat n \space ds=\iint_\limits {R}(\vec \nabla \times \vec F). \hat n \space \frac{dxdy}{\hat n.\hat k}\]
where R is the region covered by circle of radius a
\[\therefore \hat n=\frac{1}{\sqrt{2}}(\frac{x}{a}\hat i+\frac{y}{a}\hat j+\hat k)=\frac{1}{\sqrt{2}a}(x \hat i+y \hat j+a \hat k)\]
Now
\[(\vec \nabla \times \vec F).\hat n=\frac{1}{\sqrt{2}a}(y(6x^2-x)-(3y^3+y)+3ax^2)\]
Also
\[\hat n . \hat k=\frac{1}{\sqrt{2}} \implies \frac{1}{\hat n . \hat k}=\sqrt{2}\]\[(\vec \nabla \times \vec F).\hat n \frac{1}{\hat n . \hat k}=\frac{1}{a}(y(6x^2-x)-(3y^3+y)+3ax^2)\]\[\frac{(\vec \nabla \times \vec F).\hat n}{\hat n . \hat k}=\frac{1}{a}(y(6x^2-x-1)-3y^3+3ax^2)\]
Thus our surface integral is
\[\therefore \frac{1}{a}\iint_\limits R [y(6x^2-x-1)-3y^3+3ax^3]dydx\]
Using a short hand notation for the left side
\[\iint_\limits R =\frac{1}{a} \int\limits_{-a}^{a}[\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}((6x^2-x-1)y-3y^3+3ax^2)dy]dx\]
\[\iint_\limits R=\frac{1}{a}\int\limits_{-a}^{a}[(6x^2-x-1)\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}ydy-3\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}y^3dy+3ax^2\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}dy]dx\]
\[\iint_\limits R=\frac{1}{a}\int\limits_{-a}^{a}[0+0+6ax^2\int\limits_{0}^{\sqrt{a^2-x^2}}dy]dx=\frac{1}{a}\int\limits_{-a}^{a}6ax^2\sqrt{a^2-x^2}dx\]
\[\iint_{R}=6\int\limits_{-a}^{a}x^2{\sqrt{a^2-x^2}}dx=12\int\limits_{0}^{a}x^2\sqrt{a^2-x^2}dx\]\[x=a\sin(\theta)\]\[\therefore dx=a\cos(\theta)d \theta\]
The limits transform as
\[0 \rightarrow 0 \space , \space a \rightarrow \frac{\pi}{2}\]
\[\iint_\limits R=12 \int\limits_{0}^{\frac{\pi}{2}}a^2\sin^2(\theta).a^2\cos^2(\theta)d \theta=12a^4\int\limits_{0}^{\frac{\pi}{2}}(\sin(\theta)\cos(\theta))^2d \theta\]
\[\iint_\limits R=3a^4 \int\limits_{0}^{\frac{\pi}{2}}(2\sin(\theta)\cos(\theta))^2d \theta=3a^4\int\limits_{0}^{\frac{\pi}{2}}\sin^2(2 \theta)d \theta\]\[\iint_\limits R=\frac{3a^4}{2}\int\limits_{0}^{\frac{\pi}{2}}(1-\cos(4\theta))d \theta=\frac{3a^4}{2}[\theta-\frac{\sin(4 \theta)}{4}]_{0}^{\frac{\pi}{2}}\]\[\iint_\limits R=\frac{3a^4}{2}[\frac{\pi}{2}-\frac{\sin(2\pi)}{4}-(0+\frac{\sin(0)}{4})]=\frac{3a^4}{2}(\frac{\pi}{2}+0+0+0)=\frac{3 \pi a^4}{4}\]