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Computing $$I=\iint_S (y+z)\mathrm{d}y\mathrm{d}z+(z+x)\mathrm{d}z\mathrm{d}x+(x+y)\mathrm{d}x\mathrm{d}y,$$ where $S$ is the upper side of the plane $x+y+z=1$ located inside the interior of the sphere $x^2+y^2+z^2 \leq 1$.

I have tried two methods: First, I attempted to close the surface and use Divergence Theorem, but the integral over the sphere surface was also difficult to calculate. Second, I directly projected the integral onto the $xy$-plane, but the projected surface became an ellipse, and I am not familiar with handling such cases.

I wonder if there are any other solutions available.

I found a stupid mistake in my second solution. I've corrected it and wrote an answer.

MKCCT
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  • The ellipse sounds promising, could you include your work up until that? – Lorago Jun 22 '23 at 15:53
  • What do you mean by "the upper side of the plane"? Isn't $S$ just the intersection of that plane with the unit ball? – Anne Bauval Jun 22 '23 at 15:56
  • @Lorago Yes, I think it was $\iint_D (1-x-y) \mathrm{d}x\mathrm{d}y$ where $D$ is $x^2+y^2+xy-x-y \leq 0$. – MKCCT Jun 22 '23 at 15:59
  • Could you edit it into your question? – Lorago Jun 22 '23 at 15:59
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    A bit more details would be appreciated. Would computing that integral over the ellipse answer your question? Or is there more you need to compute at that point? – Lorago Jun 22 '23 at 16:03
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    @Anne Bauval the plane has orientation. – MKCCT Jun 22 '23 at 16:04
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    Since $\nabla\cdot(y+z, x+z, x+y) = 0$, you can use Stokes' theorem to turn this into a line integral of some vector field around the boundary of $S$. – eyeballfrog Jun 22 '23 at 16:14
  • Probably far more work than is required, but you could rotate the system by $\sec^{-1}\sqrt3$ about the vector $\pm\langle-1,1,0\rangle$ to make the plane parallel to $z=0$, so the projection is a circle centered at the origin. – user170231 Jun 22 '23 at 16:24
  • To finish off @user170231's idea, you can get the radius and center of the circle by using this answer, then just apply the transformation to the center and you have your new circle to integrate over – Lorago Jun 22 '23 at 16:27
  • @Lorago I've added more details about my approach. – MKCCT Jun 22 '23 at 16:27
  • Change variables in your last integral to get it to an integral over a circle, then parameterize. Pretty sure that should do the trick – Lorago Jun 22 '23 at 16:29
  • The problem with your projection method is that you use $\hat{\mathbf{n}} = (1,1,1)$ in the first line and $\mathbf{n} = (-1,-1,1)$ in the fourth instead of the correct $\mathbf{n} = (1,1,1)$. Fix these and you'll get an integral of a constant over $D$. – eyeballfrog Jun 22 '23 at 17:50
  • @eyeballfrog Yes, I also noticed that. Thank you for pointing it out. – MKCCT Jun 23 '23 at 15:20
  • @MKCCT I replaced my (incorrect) answer with one that just uses the divergence theorem and formulas for areas and volumes of well-known things. – Zarrax Jun 23 '23 at 19:50

5 Answers5

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The problem is greatly simplified by noting that the vector field $\mathbf{F} =(y+z)\hat{\mathbf{x}} + (x+z)\hat{\mathbf{y}} + (x+y)\hat{\mathbf{z}}$ can also be written as $\mathbf{F} = (x+y+z)(\hat{\mathbf{x}}+ \hat{\mathbf{y}}+ \hat{\mathbf{z}}) - (x\hat{\mathbf{x}} + y\hat{\mathbf{y}}+ z\hat{\mathbf{z}})$. In terms of the normal vector to the plane $\mathbf{n} = (1,1,1)$, this gives $\mathbf{F} = (\mathbf{n}\cdot\mathbf{x})\mathbf{n} - \mathbf{x}$. Since the original integral is equal to $\iint_S \mathbf{F}\cdot\mathbf{n}\, dA$, we have $$ \iint_S\mathbf{F}\cdot\mathbf{n}\, dA = \iint_S[(\mathbf{n}\cdot\mathbf{x})\mathbf{n} - \mathbf{x}]\cdot\mathbf{n}\,dA = (n^2-1)\iint_S\mathbf{x}\cdot\mathbf{n}\,dA = 2\iint_SdA, $$ where for the last equality we have used $\mathbf{n}\cdot\mathbf{x} = 1$ from the definition of $S$ and $|(1,1,1)|^2 = 3$. The remaining integral is just the area of $S$, which is a circle of radius $\sqrt{2/3}$. Thus we have $\iint_S\mathbf{F}\cdot\mathbf{n}\, dA = 4\pi/3$.

eyeballfrog
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  • It should be $\mathbf{\hat{n}}\mathrm{d}A$, so the answer should be divided by $\sqrt{3}$. – MKCCT Jun 23 '23 at 15:14
  • @MKCCT No, the integral you posted is $\mathbf{F}\cdot\mathbf{n}dA$. If it were $\mathbf{F}\cdot\hat{\mathbf{n}}dA$, there would have been a $1/\sqrt{3}$ in each integrand. – eyeballfrog Jun 23 '23 at 15:30
  • If it were $\mathbf{F}\cdot\mathbf{n}\mathrm{d}A$, there would have been $(y+z)\mathrm{d}A+(x+z)\mathrm{d}A+(x+y)\mathrm{d}A$, but $\mathrm{d}A\neq \mathrm{d}x\mathrm{d}y$, etc. – MKCCT Jun 23 '23 at 16:00
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Due to symmetry, $I=3\int_S(x+y)dxdy=3\int_D(x+y)dxdy$ where D is the elliptic disk $$x^2+y^2+xy-x-y\le 0,$$ the projection of $S$ on $xy$-plane. OP also found it. $D$ is parametrized by $-\pi/4\leq\theta\leq 3\pi/4$, $0\leq r\leq r(\theta)=\frac{\sin\theta+\cos\theta}{1+\sin\theta\cos\theta}$. Hence, $$I=3\int_{-\pi/4}^{3\pi/4}\int_0^{r(\theta)}r^2(\sin\theta+\cos\theta)drd\theta =\int_{-\pi/4}^{3\pi/4}\frac{(sin\theta+cos\theta)^4}{(1+\sin\theta\cos\theta)^3}d\theta=\frac{4\pi}{3\sqrt3}.$$

Bob Dobbs
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1

Project the integral onto the $xy$-plane:

$$\begin{aligned} I &= \iint_S (y+z(x,y),z(x,y)+x,x+y)\cdot \mathrm{d}S\mathbf{\hat{n}} \\ &= \iint_D (y+z(x,y),z(x,y)+x,x+y)\cdot \mathbf{\hat{n}}\sqrt{1+z_x^2+z_y^2}\mathrm{d}x\mathrm{d}y \\ &= \iint_D (1-x,1-y,x+y)\cdot \mathbf{n}\mathrm{d}x\mathrm{d}y \\ &= \iint_D (1-x,1-y,x+y)\cdot (1,1,1)\mathrm{d}x\mathrm{d}y \\ &= 2\iint_D\mathrm{d}x\mathrm{d}y = 2S_D, \end{aligned}$$ where $D$ is the projection of $x+y+z=1 \land x^2+y^2+z^2\leq1$ (a circle of radius $\sqrt{2/3}$). Since the cosine of the angle between $x+y+z=1$ and $z=0$ is $1/\sqrt{3}$, the final answer is $2\pi(2/3)\cdot (1/\sqrt{3})=\dfrac{4\pi}{3\sqrt{3}}$.

MKCCT
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Take $$\omega=(xy-xz)dx+(yz-yx)dy+(zx-zy)dz,$$Note that $$d\omega=d\left((xy-xz)dx+(yz-yx)dy+(zx-zy)dz\right)=(y+z)dydz+(x+z)dy+(x+y)dxdy,$$ so you can use Stoke's Theorem to compute the line integral of $\omega$ over your boundary, which is a circle.

Guillermo García Sáez
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  • I'm not sure whether I understand your meaning. If we use Stokes' Theorem to calculate that line integral, we would be back to the starting point of the problem. – MKCCT Jun 22 '23 at 16:48
  • It's the Stokes theorem by backwards, not as a volumen integral, a line integral since $\omega$ is a primitive of the $2$-form given to you – Guillermo García Sáez Jun 22 '23 at 16:49
  • So did you mean computing the line intergral directly? But it seems that the circle does not have a simple parametric equation... – MKCCT Jun 22 '23 at 16:56
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Since you're on the plane $x + y + z = 1$, the integral is the same as $$I=\iint_S (1 - x)\mathrm{d}y\mathrm{d}z+ (1- y) \mathrm{d}z\mathrm{d}x+(1 - z)\mathrm{d}x\mathrm{d}y,$$

Since $\iint_S \mathrm{d}y\mathrm{d}z+ \mathrm{d}z\mathrm{d}x+\mathrm{d}x\mathrm{d}y$ equals ${\sqrt 3}$ times the area of the circle, which has radius $\sqrt{2 \over 3}$, we get a contribution of ${2 \pi \over \sqrt{3}}$ from this term. So the answer will be ${2 \pi \over \sqrt{3}}$ minus $$J=\iint_S x\,\mathrm{d}y\mathrm{d}z+ y \,\mathrm{d}z\mathrm{d}x+z\,\mathrm{d}x\mathrm{d}y,$$ The integral $J$ is amenable to the divergence theorem applied to the cone whose vertex is the origin and whose base is $S$. The triple integral coming from the divergence theorem will equal $3$ times the volume of this cone, and the double integral from the conical surface will equal zero since $\langle x,y,z \rangle$ is parallel to the surface.

The volume of the cone readily computed using the fact that the circle is at distance ${1 \over \sqrt{3}}$ from the center of the sphere and has radius $\sqrt{2 \over 3}$. We obtain ${2\pi \over 9} {\sqrt{1 \over 3}}$. Three times this is ${2\pi \over 3 \sqrt{3}}$.

So the overall integral is ${\displaystyle {2 \pi \over \sqrt{3}} - {2\pi \over 3 \sqrt{3}} = {4\pi \over 3\sqrt{3}}}$.

Zarrax
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