Let $\mathcal S$ the surface of cartesian equation $z=f(x,y)=x^2-y^2$ and $$V=\left\{(x,y,z)\in \mathbb R^3 : x^2+y^2<4\right\}$$

Write the parametric equation of the portion of surface obtained by the intersection $\mathcal S\cap V$, and the cartesian equation of the tangent plane in the point $P(1,-1,0)$. Then compute $$\int\limits_{\mathcal S\cap V}\frac{z+y^2}{\sqrt[]{1+4x^2+4y^2}}\mathrm d S$$

My attempt

I used cylindrical coordinates: $$\mathcal S\cap V:\begin{cases}x=\rho\cos\theta\\y=\rho\sin\theta\\z=\rho^2\cos(2\theta)\end{cases}\quad 0\le\rho\le2$$

Now I compute the partial derivatives and I evaluate them in $P(1,-1,0)$ which in cylindrical coordinates is $(\sqrt[]2,-\pi/4,0)$ $$\frac{\partial\mathcal S}{\partial\rho}=(\cos\theta,\sin\theta,2\rho\cos2\theta)=\left(\frac{1}{\sqrt[]2},\frac{1}{\sqrt2},0\right)\\\frac{\partial \mathcal S}{\partial \theta}=(-\rho\sin\theta,\rho\cos\theta,-2\rho^2\sin2\theta)=\left(1,1,4\right)$$ Now the cross product is $$(2 \sqrt[]2, -2 \sqrt[]2, 0)$$ Then the tangent plane should be $$2 \sqrt[]2(x-1)-2 \sqrt[]2(y+1)=0$$ Alternatively, I think I could have used the following parametrization: $$\mathcal S\cap V:\begin{cases}x=t\\y=s\\z=f(t,s)=t^2-s^2\end{cases}$$ Then the normal vector is $$(-2,-2,1)\implies \text{ tangent plane }: -2(x-1)-2(y+1)+z=0$$ Why the two planes are different?

As for the integral, $$\int\limits_{\mathcal S\cap V}\frac{z+y^2}{\sqrt[]{1+4x^2+4y^2}}\mathrm d S=\int\limits_{\mathcal S\cap V}\frac{z+y^2}{\sqrt[]{1+4x^2+4y^2}}\|N\|\mathop{\mathrm d S}$$ $$\|N\|=\sqrt{1+4x^2+4y^2}\implies \int\limits_{\mathcal S\cap V}z+y^2\mathrm d S$$ Now, with cylindrical coordinates, $$\int_0^2\mathop{\mathrm d \rho}\int_0^{2\pi}\rho(\rho^2\cos^2\theta)\mathop{\mathrm d \theta}=4\pi$$


(1) The parametrisation: The boundaries of $\rho$ should be $0\leq \rho < 2$, because the points in $V$ satisfy $x^2+y^2<4$. But this is only a small detail...

(2) The plane: I believe you are mixing up the points $(1,-1,0)$ and $(1,1,0)$ in your calculation. Evaluating the partial derivatives at the point $(1,-1,0)$ give $$ \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0\right) \quad \text{and}\quad (1,1,4). $$ The cross product of these is $$ \sqrt{2}(-2,-2,1). $$ This means that the equation of the plane is $-2(x-1)-2(y+1)+z=0$.

(3) The integral: The notation looks a bit dodgy and you do calculations with both cartesian cylindrical coordinates. I suggest to work only with the parametrisation. Also, the integrand at the end seems incorrect.

First note that $N(\rho,\theta)=(-2 \rho^2 \cos\theta, 2 \rho^2 \sin t, \rho)$, so $\|N(\rho,\theta)\|=\rho \sqrt{1+4\rho^2}$. Then, substituting the parametrisation in the integral gives $$ \begin{align*} \int_{\mathcal{S}\cap V} \frac{z+y^2}{\sqrt{1+4x^2+4y^2}}\,dS &= \int_0^2 \int_0^{2\pi} \frac{\rho^2\cos 2\theta +\rho^2 \sin^2\theta}{\rho\sqrt{1+4\rho^2}} \|N(\rho,\theta)\|\, d\theta d\rho \\ &= \int_0^2 \int_0^{2\pi} \rho^2(\cos 2\theta + \sin^2\theta)\, d\theta d\rho \\ &= \left(\int_0^2 \rho^2\,d\rho \right)\left( \int_0^{2\pi}\cos^2\theta\,d\theta\right) \\ &= \frac{8}{3}\pi. \end{align*} $$

  • $\begingroup$ (1, 2) Ok I understood, I made a mistake when computing $\sin(1/\sqrt 2)$ (3) Don't you need the Jacobian determinant for the integral? I added a $\rho$ in the integrand for that purpose $\endgroup$
    – Lorenzo B.
    Nov 4 '18 at 18:25
  • 1
    $\begingroup$ About your question: the norm $\|N\|$ acts here as the jacobian for the integral. So you should not add the $\rho$. $\endgroup$
    – Ernie060
    Nov 4 '18 at 19:57

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