Question

[Calculus 3] Use a double integral to find the area of the surface over region R.
z=2x+2y+8; R = {(x,y) x^2+y^2<=4

Answer 1

Here is my work and I was hoping someone could confirm or correct it because I'm not sure if I did this right.
Since it was a circle I converted to polar so I have \[r=\sqrt{x^2+y^2}\]

Answer 2

I also have \[z=2\cos \theta + 2\sin \theta + 8\]

Answer 3

My double integral that I set up is as follows
\[4\int\limits_{0}^{\pi/2}\int\limits_{0}^{2}r(2\cos \theta + 2\sin \theta +8)drd \theta\]

Answer 4

I'm not sure if my thinking is correct but the reason I put \[4\int\limits_{0}^{\pi/2}\] is because I believe we are integrating inside the entire circle which is 0 to 2pi and since this is a symmetrical region I did 4 times the integral of 1/4 of the unit circle.

Answer 5

\[\int\limits_{0}^{2}r(2\cos \theta +2\sin \theta +8)dr = \frac{ r^2 }{ 2 }(2\cos \theta +2\sin \theta + 8) [0,2]\]

Answer 6

My final answer being \[32\pi-32\]

Answer 7

@stamp

Answer 8

Yeah I just realized I may have done this problem wrong because there is a general equation to solving these problems.

Answer 9

\[S=\int\limits_{}^{}\int\limits_{}^{}_R \sqrt{1+[f_x]^2 + [f_y]^2 }dA\]

Answer 10

Now my answer would be 6pi because the problem doesn't ask for a specific quadrant so the entire region would just be the circle.

Answer 11

In rectangular coordinates, the double integral is given by
\[\large\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\sqrt{\left(2\right)^2+\left(2\right)^2+1}\;dy\;dx\]
In polar coordinates,
\[\large\int_{0}^{2\pi}\int_{0}^{2}\sqrt{\left(2\right)^2+\left(2\right)^2+1}\;r\;dr\;d\theta\]
The problem I'm seeing with your first integral set up is that it represents the volume of the solid bounded by the region and the surface.

And as for the "symmetric" reasoning: It doesn't hold up. I'll try to draw to explain:
Basically, you have a cylinder whose top is cut off with a slant, if that makes sense.