I am stuck on the following question :(

$F(x,y,z)=(y+z)i+(x+z)j+(x+z)k$. The sphere $x^2+y^2+z^2=a^2$ intersects the postive x−, y−, and z−axes at points A, B, and C, respectively. The simple closed curve K consists of the three circular arcs AB, BC, and CA. S denotes the surface ABC of the octant of the sphere bounded by K, oriented away from the origin. Let T denote the unit tangent vector to K, and n the unit normal vector to S.

Evaluate the line integral and then the surface integral. Stoke's theorem is not to be used.


For the line integral I am required to use the equation $\int F\cdot T ds$. As the question is asking for the line integral along the bounds of the first octant I will use $C1$ to represent the line from A to B, $C2$ for B to C and $C3$ for C to A.

So we wish to evaluate the line integral over the lines $C1 = a-x^2$, $C2 = a-y^2$ and $C3 = a-z^2$. Correct?

I plan to then parametrise each of the lines and solve the integral of the parametrised equation for $F$. I am not entirely sure how to go about parametrising the lines though... Do I convert $C1$, $C2$, and $C3$ to spherical coordinates first? What do I do with the $a$ in each when I parametrise?

Thanks again for your help!

  • $\begingroup$ I don't get it: on the $\;xy$-plane, for example, the arc $\;AB\;$ is parametrized as $\;y=\sqrt{a^2-x^2}\;,\;\;0\le x\le a\;$ , or if you prefer by $\;(a\cos t\,,\,a\sin t)\;,\;\;0\le t\le\frac\pi2\;$ . How did you get those $\,C_1,C_2,C_3\;$ of yours? $\endgroup$
    – DonAntonio
    Jun 2, 2014 at 5:46

1 Answer 1


Are you sure that F is exactly the same as you written it? For complete symmetry it might be $F(x,y,z)=(y+z)i+(x+z)j+(x+y)k$. However, that isn't affecting the way of solution.

For the linear integral $\oint \mathbf{F}\cdot \boldsymbol{\tau } ds = \oint \mathbf{F}\cdot \mathbf{ds} = \int_{xy}+\int_{yz}+ \int_{xz}$ the calculations are the same in all three planes. For example, in $xy$-plane: $ z = 0, y = \sqrt{a^2-x^2} $ $$ \frac{dy}{dx} = -\frac{x}{\sqrt{a^2-x^2}}\\ \mathbf{ds} =\begin{pmatrix} 1\\ -\frac{x}{\sqrt{a^2-x^2}}\\ 0 \end{pmatrix}dx\\ \int_{xy} \mathbf{F}(x, y(x), 0)\cdot \mathbf{ds} = \int_0^a(\sqrt{a^2-x^2}-\frac{x^2}{\sqrt{a^2-x^2}})dx = x \sqrt{a^2 - x^2}\Bigr{|}_0^a = 0\\ $$ $\int_{yz}=(\pi-2) \frac{a^2}{4}$ and $\int_{xz}=\pi \frac{a^2}{4}$ together they give $\oint \mathbf{F} \cdot \mathbf{ds} = a^2 \frac{\pi-1}{2}$

For the surface integral $\iint_S \mathbf{F}\cdot \mathbf{dS}$ lets go to spherical coordinates for diversity.

$$ \left\{\begin{matrix} x = \rho cos\varphi cos\theta \\ y = \rho sin\varphi cos\theta \\ z = \rho sin\theta \end{matrix}\right.\\ Jacobian\ of\ transition\ \Delta = \rho^2cos\theta\\ \mathbf{dS} =\begin{pmatrix} cos\varphi cos\theta\\ sin\varphi cos\theta\\ sin\theta \end{pmatrix}d\varphi d\theta\\ your\ \mathbf{F} = \rho\begin{pmatrix} sin\varphi cos\theta+sin\theta\\ cos\varphi cos\theta+sin\theta\\ cos\varphi cos\theta+sin\theta \end{pmatrix}d\varphi d\theta\\ \iint_S \Delta \cdot \mathbf{F} \cdot \mathbf{dS} = \int_0^{\frac{\pi}{2}}\rho^3d\theta\int_0^{\frac{\pi}{2}}(2\cdot sin\varphi cos\varphi cos^2\theta + 2 \cdot cos\varphi sin\theta cos\theta + sin\varphi sin\theta cos\theta + sin^2\theta)d\varphi=\int_0^{\frac{\pi}{2}}\rho^3(\frac {\pi}{2} sin^2\theta+cos^2\theta+3 sin\theta cos\theta) d\theta = \frac{a^3}{8} (12+2 \pi+\pi^2) $$


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