Explanation Given: F → = 〈 2 x y , x , y + z 〉 , the surface z = 1 − x 2 − y 2 for x 2 + y 2 ≤ 1 . Formula used: Stokes’ theorem is given by- ∮ c F → . d r → = ∬ s ( ∇ → × F → ) . d s → Where, d r → is the length element on the closed curve c d s → is the surface normal to the surface s enclosed by curve c From the given data we know that F → = 〈 2 x y , x , y + z 〉 and the surface z = 1 − x 2 − y 2 for x 2 + y 2 ≤ 1 . Boundary of the surface can be defined by x 2 + y 2 = 1 i.e. r → ( t ) = 〈 cos ( t ) , sin ( t ) , 0 〉 where 0 ≤ t ≤ 2 π d r → = 〈 − sin ( t ) , cos ( t ) , 0 〉 d t Thus, ∮ F → . d r → = ∫ 0 2 π 〈 2 x y , x , y + z 〉 . 〈 − sin ( t ) , cos ( t ) , 0 〉 d t = ∫ 0 2 π 〈 2 cos ( t ) sin ( t ) , cos ( t ) sin ( t ) + 0 〉 . 〈 − sin ( t ) , cos ( t ) , 0 〉 d t = ∫ 0 2 π ( − 2 cos ( t ) sin 2 ( t ) + cos 2 ( t ) ) d t = − 2 ∫ 0 2 π ( sin 2 ( t ) d ( sin

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Chapter 18.2, Problem 2E

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The validity of Stokes’ theorem for the given vector field and surface, oriented with an upward-pointing normal.

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Chapter 18.2, Problem 1E

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The validity of Stokes’ theorem for the given vector field and surface, oriented with an upward-pointing normal.

Solution Summary: The author explains how Stokes' theorem is verified for the given vector field and surface, oriented with an upward-pointing normal.

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The validity of Stokes’ theorem for the given vector field and surface, oriented with an upward-pointing normal.

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