double integrals in polar coordinates

[cheffy]

Find the area of the region enclosed by the positive x-axis and spiral

\(\displaystyle \
r = \frac{{4\vartheta }}{3}
\\),
\(\displaystyle \
0 \le \vartheta \le 2\pi
\\)

The region looks like a snail shell.

I'm getting confused as to where my bounds should be. Would it be 0<r<8pi/3 and 0<theta<pi/2?

 

[Deleted member 4993]

Find the area of the region enclosed by the positive x-axis and spiral
\(\displaystyle \
r = \frac{{4\vartheta }}{3}
\\),
\(\displaystyle \
0 \le \vartheta \le 2\pi
\\)
The region looks like a snail shell.

I'm getting confused as to where my bounds should be. Would it be 0<r<8pi/3 and 0<theta<pi/2?

I assume you plotted the function.

This one I think is called "Archemedsis' spiral".

Since there is no symmetry to help us - theta should be from 0 to 2[pi]

 

[cheffy]

Find the area of the region enclosed by the positive x-axis and spiral
\(\displaystyle \
r = \frac{{4\vartheta }}{3}
\\),
\(\displaystyle \
0 \le \vartheta \le 2\pi
\\)
The region looks like a snail shell.

I'm getting confused as to where my bounds should be. Would it be 0<r<8pi/3 and 0<theta<pi/2?

I assume you plotted the function.

This one I think is called "Archemedsis' spiral".

Since there is no symmetry to help us - theta should be from 0 to 2[pi]

But then how would I incorporate the fact that they only want it above the positive x-axis?

 

[galactus]

Here's a graph of your Archimedean spiral:

For the area above the x-axis, it should be:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\pi}\left[\frac{4{\theta}}{3}\right]^{2}d{\theta}\)

 

[cheffy]

Here's a graph of your Archimedean spiral:

For the area above the x-axis, it should be:

\(\displaystyle \L\\\frac{1}{2}\int_{0}^{\pi}\left[\frac{4{\theta}}{3}\right]^{2}d{\theta}\)

But that's not a double integral. Where's the r?

And the question says the positive x-axis, which I assume means only the x-axis to the right of the y-axis. Wouldn't it be pi/2?

 

[galactus]

OK, sorry, I didn't pay close enough attention. You must have a double integral, huh?.

Since we're only in the first quadrant, yes, it would be Pi/2.

Then r would vary from 0 to 2Pi/3

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{3}}{\frac{4{\theta}}{3}r}drd{\theta}\)

Here's a 3D, for kicks.

 

[cheffy]

OK, sorry, I didn't pay close enough attention. You must have a double integral, huh?.

Since we're only in the first quadrant, yes, it would be Pi/2.

Then r would vary from 0 to 2Pi/3

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{3}}{\frac{4{\theta}}{3}r}drd{\theta}\)

Here's a 3D, for kicks.

Where did that r come into the integral from? Do I automatically include it with drdtheta?

I thought it would just be \(\displaystyle \L\\\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{2\pi}{3}}{\frac{4{\theta}}{3}}drd{\theta}\)

 

[galactus]

That 'other r' is part of polar integration. Look it up.

\(\displaystyle \L\\\int_{\alpha}^{\beta}\int_{r_{1}({\theta})}^{r_{2}({\theta})}f(f,{\theta})rdrd{\theta}\)