Computing a surface integral with polar coordinates

[Yami]

Homework Statement

Show that $\iint_{S}(x^2 + y^2)d\sigma = \frac{9\pi}{4}$
where $S = \{(x,y,z): x > 0, y > 0, 3 > z > 0, z^2 = 3(x^2 + y^2)\}$

Homework Equations

$\iint_{S}f(x,y,z)d\sigma = \iint_{R}f(r(x,y))\sqrt{[r_x(x,y)]^2 + [r_y(x,y)]^2 + 1}$
where $r : R → ℝ^3, R \in ℝ^2$ parametrizes the surface S.

The Attempt at a Solution

$\iint_{S}(x^2 + y^2)d\sigma = \iint_{R}(x^2 + y^2)\sqrt{\frac{9^2}{3(x^2 + y^2)} + \frac{9^2}{3(x^2 + y^2)}+ 1}dxdy$
$= \int_{0}^{2\pi}\int_{0}^{3}r^2\sqrt{\frac{9r^2}{3r^2} + 1} drd\theta$ (polar coordinates)
$= \int_{0}^{2\pi}\int_{0}^{3}2r^2drd\theta = 36\pi$

This is obviously wrong. I'm guessing I have the wrong domain restrictions
$R = \{ (r, \theta) : 0 < \theta < 2\pi, 0 < r < 3\}$
If so, I can't figure out how to get the correct restriction.

 

[DryRun]

You should draw the graph to find the limits. S is the surface of a cone in the first octant.
$$z=\pm \sqrt{3x^2+3y^2}$$
But since we are considering the cone in the 1st octant, where $z≥0$, thus:
$$z=\sqrt{3x^2+3y^2}$$
Find $z_x$ and $z_y$ as you'll be projecting the surface S onto the x-y plane.

The surface area in terms of Cartesian coordinates:
$$\iint_{S}(x^2 + y^2)d\sigma = \iint_{Ω} (x^2 + y^2) \sqrt {(z_x)^2+(z_y)^2+1} \,.dxdy=\iint_{Ω} (x^2 + y^2) \sqrt {\frac{12x^2+12y^2}{3x^2+3y^2}} \,.dxdy=2\iint_{Ω} (x^2 + y^2) \,.dxdy$$ where Ω is the projection of S onto the x-y plane.

To find the equation of the points of intersection of the plane z = 3 and the cone, substitute z = 3 into the equation of the cone.
$$3^2=3x^2+3y^2<br /> \\x^2+y^2=3<br /> \\r^2=3$$
The projected area in terms of polar coordinates:
$$2 \iint_Ω r^2 \,.rdrd\theta$$
Now, using the graph of the region Ω, find the limits. See attached graph.

But from the projection onto the x-y plane, you only need a quarter of Ω (since the section of surface S required is found in the 1st octant), so $0\le \theta \le \pi /2$

 

[Yami]

Alright, I got it. Thanks for your help.
I think I get the gist of the radians restriction; so if we have just x > 0, y > 0, but no z > 0 would it be $R = \{(r, \theta); 0 < r < \sqrt{3}, 0 < \theta < \pi\}$?

 

[DryRun]

Alright, I got it. Thanks for your help.
I think I get the gist of the radians restriction; so if we have just x > 0, y > 0, but no z > 0 would it be $R = \{(r, \theta); 0 < r < \sqrt{3}, 0 < \theta < \pi\}$?

You are correct except for the limits of $\theta$, which should be:
$$0 < \theta < \pi /2$$ as the surface area that this problem requires is found in the first octant, or in case of no z-axis, then it's the 1st quadrant.

 

[Yami]

Okay so if it's just {y > 0, but no x > 0, z > 0} OR {x > 0, but no y > 0, z > 0}, two quadrants of the space, then would it be $0 < \theta < \pi$?

 

[DryRun]

Okay so if it's just {y > 0, but no x > 0, z > 0} OR {x > 0, but no y > 0, z > 0}, two quadrants of the space, then would it be $0 < \theta < \pi$?

Draw the Cartesian axes in a 3D coordinate system; x, y and z, perpendicular to each other. Label them with simple values {-1,0,1} for each axis. For example, if x > 0, then it means you are only considering the positive x axis. As you are given more limits, you will be able to progressively pinpoint the exact octant required.

Likewise, if y > 0, it means the graph is along the positive y axis. You can apply this simple means of knowing the required section of graph by determining which octant or quadrant is related to the problem.

In your original problem, {x > 0, y > 0, 0 < z < 3}. Consider x > 0 and y > 0. This automatically places the required section of the graph in the first quadrant, but since your problem involves 3D coordinates, you now have an additional upper and lower dimension (the z axis). If z > 0, then the section is found in the 1st quadrant but above the z axis, meaning in the positive (upper) direction of the axis, in other words, the 1st octant. If z < 0, then it's obviously below the x-y plane, meaning, the required section of the graph is along the negative z axis.