Computing $\iint_S (y+z)\mathrm{d}y\mathrm{d}z+(z+x)\mathrm{d}z\mathrm{d}x+(x+y)\mathrm{d}x\mathrm{d}y$

Question

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.

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

Answer 1

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.

Answer 2

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 $\iint_S\mathbf{F}\cdot\mathbf{n}\, dA = 4\pi/3$.

Answer 3

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, $$I=3\int_{-\pi/4}^{3\pi/4}\int_0^{r(\theta)}r^2(\sin\theta+\cos\theta)drd\theta =\int_{-\pi/4}^{3\pi/4}\frac{(sin\theta+cos\theta)^4}{(1+\sin\theta\cos\theta)^3}d\theta=\frac{4\pi}{3\sqrt3}.$$

Answer 4

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 $2\pi(2/3)\cdot (1/\sqrt{3})=\dfrac{4\pi}{3\sqrt{3}}$.

Answer 5

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 ${\displaystyle {2 \pi \over \sqrt{3}} - {2\pi \over 3 \sqrt{3}} = {4\pi \over 3\sqrt{3}}}$.