Evaluating the surface integral of a paraboloid $z = x^2 + y^2$ for $0\le z \le 4$

Question

Evaluate $\iint_S x^2 dS$ where S is the paraboloid $z=x^2 + y^2$ for $0\le z \le 4$.

So far, I have parametrized my function as: $r(x,y) = \langle x, y, x^2 + y^2\rangle$ which gives me the partial sums: $$ r_x = \langle 1, 0, 2x\rangle$$ $$ r_y = \langle 0, 1, 2y\rangle$$ which gives me $ r_x \times r_y = \langle-2x, -2y, 1\rangle $. I'm a bit confused which orientation should I use and what my surface integral will be.

Answer 1

Rather than keeping the variables $x$ and $y$, consider parameterizing $S$ in cylindrical coordinates as

$$\vec s(r,\theta)=\langle r\cos\theta,r\sin\theta,r^2\rangle$$

which follows from the change of coordinates,

$$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=x^2+y^2\end{cases}$$

where $0\le r\le2$ and $0\le\theta\le2\pi$. (Keeping the parameterization in terms of $x$ and $y$ would make the limits of integration more annoying to deal with.)

Now let the normal vector (again, direction doesn't matter) be

$$\vec n=\vec s_\theta \times \vec s_r=\langle 2r^2\cos t,2r^2 \sin t,-r\rangle$$

which makes the surface element

$$\mathrm dS=\|\vec n\|\,\mathrm dr\,\mathrm d\theta=r\sqrt{4r^2+1}\,\mathrm dr\,\mathrm d\theta$$

So the surface integral is

$$\begin{align} \iint_S x^2\,\mathrm dS&=\int_0^{2\pi}\int_0^2 (r\cos\theta)^2 r\sqrt{4r^2+1}\,\mathrm dr\,\mathrm d\theta\\[1ex] &=\left(\int_0^{2\pi}\cos^2\theta\,\mathrm d\theta\right)\left(\int_0^2 \, r^3 \sqrt{4r^2+1}\,\mathrm dr\right) \end{align}$$

which is easy to compute.

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If you want to hold on to your original parameterization, you would have

$$\vec r(x,y)=\langle x,y,x^2+y^2\rangle$$

with $-2\le x\le2$ and $-\sqrt{4-x^2}\le y\le\sqrt{4-x^2}$. Then the normal vector would be as you found,

$$\vec n=\vec r_x\times\vec r_y=\langle-2x,-2y,1\rangle$$

so that the surface element is

$$\mathrm dS=\|\vec n\|\,\mathrm dx\,\mathrm dy=\sqrt{4x^2+4y^2+1}\,\mathrm dx\,\mathrm dy$$

and the integral would be

$$\iint_S x^2\,\mathrm dS=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}x^2\sqrt{4x^2+4y^2+1}\,\mathrm dy\,\mathrm dx$$

Answer 2

Your integrand is a scalar. As the surface is defined by $z = x^2 + y^2$, you can apply the surface area formula directly to map $dS$ to $dA$ which is projection in $XY$ plane.

$dS = \sqrt{1 + \big(\frac{\partial z}{\partial x} \big)^2 + \big(\frac{\partial z}{\partial y} \big)^2} \ dA$

Now, $\sqrt{1 + \big(\frac{\partial z}{\partial x} \big)^2 + \big(\frac{\partial z}{\partial y} \big)^2} = \sqrt{1 + 4x^2+4y^2}$

Converting it into polar coordinates, $x = r \cos \theta, y = r \sin \theta$,

$0 \leq z \leq 4 \implies 0 \leq r \leq 2$

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^2 r^3 \cos^2\theta \ \sqrt{1 + 4r^2} \ dr \ d\theta$