The integral a ≡∫S da (1.106) is sometimes called the vector area of the surface S. If S happens to be flat, then |a| is the ordinary (scalar) area, obviously. (a) Find the vector area of a hemispherical bowl of radius R. (b) Show that a = 0 for any closed surface. [Hint: Use Prob. 1.61a.] (c) Show

Question

The integral

[latex]a ≡\int_{S}^{}{} da[/latex] (1.106)

is sometimes called the vector area of the surface S. If S happens to be flat, then |a| is the ordinary (scalar) area, obviously.

(a) Find the vector area of a hemispherical bowl of radius R.

(b) Show that a = 0 for any closed surface. [Hint: Use Prob. 1.61a.]

(c) Show that a is the same for all surfaces sharing the same boundary.

(d) Show that

[latex]a =\frac{1}{2}\oint{} r × dl, [/latex] (1.107)

where the integral is around the boundary line. [Hint: One way to do it is to draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite side dl, and exploit the geometrical interpretation of the cross product (Fig. 1.8).]

(e) Show that

∮(c · r) dl = a × c, (1.108)

for any constant vector c. [Hint: Let T = c · r in Prob. 1.61e.]

Accepted Answer

(a) [latex]da = R^2\sin heta d heta d\phi \hat{r}[/latex] . Let the surface be the northern hemisphere. The [latex]\hat{x} and \hat{y}[/latex] components clearly integrate to zero, and the [latex]\hat{z}[/latex] component of [latex]\hat{r}[/latex] is cos θ, so

[latex]a =\int R^2 \sin heta \cos heta d heta d\phi \hat{z} =2\pi R^2\hat{z} \int_{0}^{\pi /2}{}\sin heta \cos heta d heta =2\pi R^2 \hat{z}\frac{\sin ^2 heta }{2}\mid ^{\pi /2}_{0}=\pi R^2 \hat{z}[/latex]

(b) Let T = 1 in Prob. 1.61(a). Then ∇T = 0, so ∮ da = 0. qed

(c) This follows from (b). For suppose [latex]a_1 ≠ a_2;[/latex] then if you put them together to make a closed surface, [latex]∮da = a_1 − a_2≠0[/latex] .

(d) For one such triangle, [latex]da =\frac{1}{2}\left(r imes dl ight) [/latex] (since r × dl is the area of the parallelogram, and the direction is perpendicular to the surface), so for the entire conical surface, [latex]a =\frac{1}{2}\oint{}r imes dl. [/latex] .

(e) Let T = c · r, and use product rule #4: ∇T = ∇(c · r) = c × (∇×r) + (c · ∇)r. But ∇×r = 0, and (c ·∇)r = [latex]\left(c_x\frac{\partial}{\partial x} +c_y\frac{\partial}{\partial y}+c_z\frac{\partial}{\partial z} ight) (x \hat{x} + y \hat{y} + z \hat{z} ) =c_x \hat{x} + c_y \hat{y} + c_z \hat{z} = c.[/latex] So Prob. 1.61(e) says

∮T dl =∮(c · r) dl = −∫(∇T) × da = −∫c× da = −c ×∫da = −c × a = a × c. qed

Question 1.62: The integral a ≡∫S da (1.106) is sometimes called the vector...

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