March 20th, 2019, 08:24 AM  #1 
Member Joined: Apr 2017 From: India Posts: 56 Thanks: 0  Stokes Theorem
I am unable to solve this question of Stokes Theorem concept. It seems tough to be due to the presence of intersection of plane and sphere. Please help with the correct option and show the procedure. I am thankful in advance.

March 20th, 2019, 09:11 AM  #2 
Newbie Joined: Oct 2018 From: USA Posts: 19 Thanks: 13 Math Focus: Algebraic Geometry 
Stokes says that: $\displaystyle \int_C \vec{F}\cdot d\vec{r} = \iint_{D} curl(F) \cdot ndA$ In this case, $\vec{F} = yi+zj+xk$ and $curl(F) = ijk$. Now, since the intersection of the the sphere and plane is a circle on the plane, the circle shares a normal with the plane. This normal is the unit gradient of the plane. $grad(f) = i+j+k$, divide by the magnitude to get the unit vector: $\displaystyle n=\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1} {\sqrt{3}}k$ Now take the dot product of the curl and the normal to receive $\sqrt{3}$ Lastly, integrate: $\displaystyle \iint \sqrt{3}dA = \sqrt{3}\iint dA$ Since the plane slices the sphere through its center, the intersection is a circle of radius $a$, therefore: $\displaystyle \sqrt{3}\iint dA = \sqrt{3} \cdot \pi a^{2}$ Now take the absolute value as the question asks to receive choice b) 

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