Verify Stokes' Theorem If$\sigma$ Is the Portion of the Sphere X² + Y²

Question

Examlex · Accepted Answer

The Answer is If

\sigma

is oriented by upward normals, then C is the intersection of the sphere x² + y² + z² = 1 with z = 0, thus, C is the circle x² + y² = 1 which can be parameterized as r(t) = cos t i + sin t j for 0

\le

t

\le

2

\pi

, so, $If \sigma is oriented by upward normals, then C is the intersection of the sphere x2 + y2 + z2 = 1 with z = 0, thus, C is the circle x2 + y2 = 1 which can be parameterized as r(t) = cos t i + sin t j for 0 \le t \le 2 \pi , so, ;Curl F = k, , and R is the circular region in the xy-plane enclosed by C, so area of the circle of radius 1 = \pi (1)2 = \pi .$ ;Curl F = k, $If \sigma is oriented by upward normals, then C is the intersection of the sphere x2 + y2 + z2 = 1 with z = 0, thus, C is the circle x2 + y2 = 1 which can be parameterized as r(t) = cos t i + sin t j for 0 \le t \le 2 \pi , so, ;Curl F = k, , and R is the circular region in the xy-plane enclosed by C, so area of the circle of radius 1 = \pi (1)2 = \pi .$ , and R is the circular region in the xy-plane enclosed by C, so $If \sigma is oriented by upward normals, then C is the intersection of the sphere x2 + y2 + z2 = 1 with z = 0, thus, C is the circle x2 + y2 = 1 which can be parameterized as r(t) = cos t i + sin t j for 0 \le t \le 2 \pi , so, ;Curl F = k, , and R is the circular region in the xy-plane enclosed by C, so area of the circle of radius 1 = \pi (1)2 = \pi .$ area of the circle of radius 1 =

\pi

(1)² =

\pi

.

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Verify Stokes' Theorem If $\sigma$ Is the Portion of the Sphere X² + Y²

Correct Answer:
Verified