Q. 5 Suppose that S 1 is the upper ha... [FREE SOLUTION]

Chapter 14: Q. 5 (page 1140)

Suppose that S 1 is the upper half of the unit sphere, with outwards-pointing normal n1, and S 2 is a balloon-shaped surface whose boundary is the unit circle whose orientation leads to counterclockwise parametrization of the unit circle. If F(x, y ,z) is a smooth vector field defined on a region large enough to include both surfaces, what is the relationship between $鈭� {鈭�}_{s_{1}} curl F . n_{1} dS and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS ?$

Short Answer

Expert verified

$By Stokes Theorem, both integrals are equivalent to {鈭�}_{C_{}} F . dr, so 鈭� {鈭�}_{S_{1}} curl F . n_{1} dS and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS are equal to each other, that is, 鈭� {鈭�}_{s_{1}} curl F . n_{1} dS = 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS$

Step by step solution

Step 1. Given

$鈭� {鈭�}_{s_{1}} c u r l F . n_{1} d S a n d 鈭� {鈭�}_{s_{2}} c u r l F . n_{2} d S$

Step 2. Objective

$Consider the following data . The surface S_{1}, is the upper half of the unit sphere with outwards - pointing normal n_{1} . The surface S_{2}, is a balloon - shaped surface whose boundary is the unit circle and whose orientation leads to counterclockwise parametrization of the unit circle . A smooth vector field F (x, y, z) is defined on a region large enough to include both surfaces, The objective is to find the relationship between 鈭� {鈭�}_{s_{1}} curl F . n_{1} dS and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS .$

Step 3. Proof .

$Apply Stokes Theorem to find the required relationship . Stokes Theorem states that, " Let S be an oriented, smooth or piecewise - smooth surface bounded by a curve C . Suppose that n is an oriented unit normal vector of S and C has a parametrization that traverses C in the counterclockwise direction with respect to n . If a vector field F (x, y, z) = F_{1} (x, y, z) i + F_{2} (x, y, z) j + F_{3} (x, y, z) k is defined on S, then, 鈭� {鈭�}_{s_{1}} curl F (x, y, z) dS = {鈭�}_{c} F (x, y, z) dr " By Stokes' theoram, 鈭� {鈭�}_{S_{1}} curl F . n_{1} dS = {鈭�}_{C_{1}} F . dr and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS = {鈭�}_{C_{2}} F . dr . Both surfaces S, and S, are bounded by the unit circle, so the boundary curves for both surfaces are the same . That is, C_{1} = C_{2} = C . Then, above two integrals becomes, 鈭� {鈭�}_{S_{1}} curl F . n_{1} dS = {鈭�}_{C_{}} F . dr and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS = {鈭�}_{C} F . dr . implies that, 鈭� {鈭�}_{s_{1}} curl F . n_{1} dS = 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS . Hence, by Stokes Theorem, both integrals are equivalent to {鈭�}_{C_{}} F . dr, so 鈭� {鈭�}_{S_{1}} curl F . n_{1} dS and 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS are equal to each other, that is, 鈭� {鈭�}_{s_{1}} curl F . n_{1} dS = 鈭� {鈭�}_{s_{2}} curl F . n_{2} dS$

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Short Answer

Step by step solution

Step 1. Given

Step 2. Objective

Step 3. Proof .

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