Test Stokes' Theorem for the function

[stunner5000pt]

Test Stokes' Theorem for the function
$\vec{v} = (xy) \hat{x} + (2yz) \hat{y} + (3yz) \hat{z} /$

for the triangular shaded region

$$\int_{S} (\grad \times v) \dot da = \oint_{P} v \bullet dl$$

for the left hand side

$$\int_{0}^{2} \int_{0}^{2} (-2y \hat{x} - 3z \hat{y} - x \hat{z}) \hat{x} dy dz = \int_{0}^{2} \int_{0}^{2} (-2y) dy dz = \left[ \frac{-y^2} \right]_{0}^{2} \int_{0}^{2} dz = (-4)(2) = -8$$

so far so good??

ok for hte right hand side

i) x=0, z=0, $v \bullet dl = 2y dy$ integration limits 0 to 2

ii) x=0, y=2, $v \bullet dl = 0 dz$ integration from 0 to 2

i am not sute about the last one though...

am i right with i and ii though??

thank you for any advice!

 

[HallsofIvy]

Test Stokes' Theorem for the function
$\vec{v} = (xy) \hat{x} + (2yz) \hat{y} + (3yz) \hat{z} /$

for the triangular shaded region

$$\int_{S} (\grad \times v) \dot da = \oint_{P} v \bullet dl$$

for the left hand side

$$\int_{0}^{2} \int_{0}^{2} (-2y \hat{x} - 3z \hat{y} - x \hat{z}) \hat{x} dy dz = \int_{0}^{2} \int_{0}^{2} (-2y) dy dz = \left[ \frac{-y^2} \right]_{0}^{2} \int_{0}^{2} dz = (-4)(2) = -8$$

so far so good??

No, your $\nabla \vec{v}$ is wrong. The $\hat{y}$ component is 0.

ok for hte right hand side

i) x=0, z=0, $v \bullet dl = 2y dy$ integration limits 0 to 2

That should be $v \bullet dl= 2yz dy= 0$

ii) x=0, y=2, $v \bullet dl = 0 dz$ integration from 0 to 2

i am not sute about the last one though...

for the last one, you can take y= t, z= 2- t integrating from t= 0 to 2.

am i right with i and ii though??

thank you for any advice!

By the way, to get $\nabla$ use "\nabla", not "\del".