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I need to calculate the integral $$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$$ being $\Gamma=S_1\cap S_2$, given:

  • $S_1=\{(x,y,z)\in\mathbb{R}:2x+2y+z=3\}$
  • $S_2=\{(x,y,z)\in\mathbb{R}:z=9-x^2-y^2\}$

In my problem i'm asked to solve it using both direct integration and the Stokes Theorem. I've started trying the Stokes part, calculating the rotational of $F$: $$F=((y^2-z^2),(z^2-x^2),(x^2-y^2))\Longrightarrow \text{rot}(F)=-2(y+z,x+z,x+y)$$ Now, I know that the normal vector $N$ is $(2,2,1)$ (because $S_1\subset\Gamma$ and that's $S_1$'s normal vector at any point).

So now I using Stokes I have that (after simplifying) $$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz=\frac{-2}{3}\iint_S(3x+3y+4z)d\sigma$$ I'm stucked here. I don't get what $S$ am I supposed to use in the double integral (I guess I must do some variable change).

For the direct integration part, i don't know where to start. I will thank any help.

Edit: I need two different solutions, one using direct integration, and another using Stokes Theorem.

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    $\begingroup$ You can use either surface for the right hand side. Stokes' theorem tells you that you get the same value for any surface integral bounded by the curve. $\endgroup$ – Paul May 26 at 18:04
  • $\begingroup$ So i can just use for example $S_1$ (the plane)? Do I need to change variables, parameterize or something like that? Thanks for your answer! $\endgroup$ – Alejandro Bergasa Alonso May 26 at 18:12
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    $\begingroup$ For the direct part, you start by determining parametric equations for $\Gamma$. $\endgroup$ – saulspatz May 26 at 18:15
  • $\begingroup$ @Paul What would the integration limits be? I don't understand how to find them. $\endgroup$ – Alejandro Bergasa Alonso May 26 at 19:22
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    $\begingroup$ @AlejandroBergasaAlonso I would use $x$ and $y$ for the parameterization. You can eliminate $z$ with the equation for the plane. Note that the limits are given by the intersection, which can be found with the equation $2x+2y+9-x^2-y^2=3$. $\endgroup$ – Paul May 26 at 19:48
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Direct approach:

Let's study the intersection by equating the $z$:

$$3-2x-2y=9-x^2-y^2$$

$$x^2+y^2-2x-2y=9-3=6$$

$$(x-1)^2+(y-1)^2=(2\sqrt2)^2$$

Hence I would use the parametrization $x=1+2\sqrt2\cos \theta, y = 1+2\sqrt2\sin \theta, z=-1-4\sqrt2\sin \theta - 4\sqrt2 \cos \theta.$

$$\frac{dx}{d\theta} = -2\sqrt2 \sin \theta, \frac{dy}{d\theta}= 2\sqrt2\cos \theta, \frac{dz}{d\theta}= -4 \sqrt2\cos \theta + 4\sqrt2\sin \theta.$$

\begin{align} &\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\ &= \int_0^{2\pi}y^2\left( \frac{dx}{d\theta} - \frac{dz}{d\theta} \right) +z^2\left( \frac{dy}{d\theta} - \frac{dx}{d\theta} \right) + x^2 \left( \frac{dz}{d\theta} - \frac{dy}{d\theta} \right)\, d\theta \\ &= \int_0^{2\pi} \left( 1+4\sqrt2\sin \theta + 8\sin^2 \theta\right)\left( -6\sqrt2\sin \theta + 4\sqrt2\cos \theta\right) \\ &+\left( 1+32 \sin^2 \theta + 32 \cos^2 \theta + 8\sqrt2 \sin \theta + 8\sqrt2 \cos \theta + 32 \sin \theta \cos \theta\right)\left( 2\sqrt2\cos \theta + 2\sqrt2\sin \theta\right) \\ &+\left( 1+4\sqrt2\cos \theta + 8\cos^2 \theta\right)\left( -6\sqrt2\cos \theta + 4\sqrt2\sin\theta\right) \, d\theta \\ &= \int_0^{2\pi} -16 \sin^2 \theta -16 \cos^2 \theta \, d\theta \\ &=\int_0^{2\pi} -16 \, d\theta \\ &= -32 \pi \end{align}

Remark: There is no need to work out the details in the middle as we know that for $\{\sin^3 \theta, \cos^3 \theta, \sin^2 \theta \cos \theta, \sin \theta \cos^2 \theta, \sin \theta \cos \theta, \sin \theta, \cos \theta\}$, they vanish when we integrate form $0$ to $2 \pi$. We just have to keep track of $\sin^2 \theta$ and $\cos^2 \theta$.


Alternatively, we can use the Stoke's theorem,

$$curl(F)=-2(y+z, x+z, x+y)$$

The normal vector is $(2,2,1)$.

The dot product is $-2(3x+3y+4z)=-2(3x+3y+4(3-2x-2y))=-2(12-5x-5y)$.

Let's integrate using the parametrization $x=1+r\cos \theta, y = 1+r \sin \theta, 0 \le \theta \le 2\pi, 0 \le r \le 2\sqrt2$:

\begin{align} &\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\ &= -2\int_0^{2\sqrt2} \int_0^{2\pi} (12-5(1+r\cos \theta)-5(1+r\sin \theta))r\, d\theta \,dr \\ &= -2 \int_0^{2\sqrt2} \int_0^{2\pi} (2r-5r^2\cos \theta-5r^2\sin \theta)\, d\theta \,dr \\ &= -2 \int_0^{2\sqrt2} \int_0^{2\pi} (2r)\, d\theta \,dr \\ &= -4\pi \int_0^{2\sqrt2} 2r \, dr \\ &= -32 \pi \end{align}

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    $\begingroup$ I think you made a mistake in the first part of the direct integration part. I think the equation of $z$ results in $$(x-1)^2+(y-1)^2=8.$$ Please check it. $\endgroup$ – Alejandro Bergasa Alonso May 30 at 9:35
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    $\begingroup$ Thanks, let me edit it. $\endgroup$ – Siong Thye Goh May 30 at 9:37
  • $\begingroup$ I don't get the step in the direct part when you do $$\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz = \int_0^{2\pi}y^2\left( \frac{dx}{d\theta} - \frac{dz}{d\theta} \right) +z^2\left( \frac{dy}{d\theta} - \frac{dx}{d\theta} \right) + x^2 \left( \frac{dz}{d\theta} - \frac{dy}{d\theta} \right)\, d\theta.$$ Could you please explain it? Thanks for your answer! $\endgroup$ – Alejandro Bergasa Alonso May 30 at 10:23
  • $\begingroup$ We have $$\int_\Gamma (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz = \int_0^{2\pi} \left[(y^2-z^2) \frac{dx}{d\theta} + (z^2-x^2) \frac{dy}{d\theta} + (x^2-y^2) \frac{dz}{d\theta}\right] \, d\theta.$$ After which, I just collect the $y^2$ terms together, the $z^2$ terms together and $x^2$ terms together. Alternatively, you can also try to work out terms like $y^2-z^2$ directly, perhaps using the formula $y^2-z^2=(y-z)(y+z)$. $\endgroup$ – Siong Thye Goh May 30 at 10:27

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