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## Main Question or Discussion Point

Hi guys!

well! in the following question i need to verify The Stoke's Theorem:

Q: Verify Stoke's theorem for

F=6zi +(2x+y)j -xk

where "S" is the upper half of the sphere [tex]x^2+y^2+z^2=1[/tex]

bounded by a closed curve "C" [tex]x^2+y^2+z^2=1[/tex] at z=0 plane.

Now here is the stoke's theorem:

[tex]\oint F.dr = \int\int (curl of F).n ds[/tex]

OKz so when i just solved this problem, i found:

Curl of F = 7j + 2k

and [tex]n=\bigtriangledown (x^2+y^2+z^2-1)= 2xi+2yj+2zk[/tex]

and then

(curl of F). n= 14y+4z

now i put it under the integral sign like :

[tex]\int\int (14y+4z)ds[/tex]

i put here the value of z as

[tex]z=\sqrt{1-x^2-y^2}[/tex]

[tex]\int\int [(14y+4(\sqrt{1-x^2-y^2})]ds[/tex]

now what i have done is this that i take

[tex]x=r\cos\theta[/tex]

[tex]y=r\sin\theta[/tex]

[tex]ds=rdrd\theta[/tex]

and i finally put it in the integral sign

[tex]\int\int [14r\sin\theta+4(\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta})]rdrd\theta[/tex]

where [tex]0 \leq r\leq 1[/tex]

[tex]0\leq \theta \leq 2\pi [/tex]

now when i solve it

i found its answer to be

[tex]\frac{8\pi}{3}[/tex]

but when i solved the left hand side of the stoke's theorem:

i.e.

[tex]\oint F.dr[/tex]

i found it that

[tex]\oint F.dr = 2\pi[/tex]

and the result is that

the stoke's theorem is not getting verified. please tell me where im making mistake in this problem.

both sides should be equal to each other. but im not getting both of them equal to each other . please help!

thanks in advance!

well! in the following question i need to verify The Stoke's Theorem:

Q: Verify Stoke's theorem for

F=6zi +(2x+y)j -xk

where "S" is the upper half of the sphere [tex]x^2+y^2+z^2=1[/tex]

bounded by a closed curve "C" [tex]x^2+y^2+z^2=1[/tex] at z=0 plane.

Now here is the stoke's theorem:

[tex]\oint F.dr = \int\int (curl of F).n ds[/tex]

OKz so when i just solved this problem, i found:

Curl of F = 7j + 2k

and [tex]n=\bigtriangledown (x^2+y^2+z^2-1)= 2xi+2yj+2zk[/tex]

and then

(curl of F). n= 14y+4z

now i put it under the integral sign like :

[tex]\int\int (14y+4z)ds[/tex]

i put here the value of z as

[tex]z=\sqrt{1-x^2-y^2}[/tex]

[tex]\int\int [(14y+4(\sqrt{1-x^2-y^2})]ds[/tex]

now what i have done is this that i take

[tex]x=r\cos\theta[/tex]

[tex]y=r\sin\theta[/tex]

[tex]ds=rdrd\theta[/tex]

and i finally put it in the integral sign

[tex]\int\int [14r\sin\theta+4(\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta})]rdrd\theta[/tex]

where [tex]0 \leq r\leq 1[/tex]

[tex]0\leq \theta \leq 2\pi [/tex]

now when i solve it

i found its answer to be

[tex]\frac{8\pi}{3}[/tex]

but when i solved the left hand side of the stoke's theorem:

i.e.

[tex]\oint F.dr[/tex]

i found it that

[tex]\oint F.dr = 2\pi[/tex]

and the result is that

the stoke's theorem is not getting verified. please tell me where im making mistake in this problem.

both sides should be equal to each other. but im not getting both of them equal to each other . please help!

thanks in advance!

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