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March 20th, 2019, 09:24 AM   #1
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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.
Attached Images 20190320_214843.jpg (80.3 KB, 6 views) March 20th, 2019, 10:11 AM #2 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 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 $curl(F) = -i-j-k$. 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) Thanks from topsquark and shashank dwivedi Tags stokes, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Robotboyx9 Calculus 1 October 16th, 2017 07:33 AM matteamanda Calculus 1 March 10th, 2017 04:50 AM djcan80 Calculus 2 August 24th, 2016 08:28 PM ZardoZ Applied Math 2 June 21st, 2011 06:31 AM george gill Calculus 5 May 14th, 2011 03:13 PM

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