HELP! Conceptual question

[Xibalba]

Hello again all,

I am reading through my lecture notes (my final's coming up) and I have some questions regarding polar coordinates. I was wondering why the extra 'r' term is in there. Does it have something to do with u-substitution? I am also having trouble figuring out the region R. For example:

Evaluate the double integral of (3x+4y^2) where R is the region in the upper half-plane bounded by the circles x^2 = y^2 = 1 and x^2 + y^2 = 4. How would I find 'R'? And where the heck does the extra r term come from?

Thanks.

 

[Deleted member 4993]

I am guessing you are talking about - elemental area dA, where:

dA = r d[theta] * dr

That leading 'r' comes from the definition 'angle' (S = r * theta) and hence ds = r*d[theta]

In your question, the bounds (limits) on 'r' is 1 to 2.(Why?)

Read the derivation part in your text book carefully.

 

[Xibalba]

Thanks so much. I will review that section in my book and come back to this afterwards.

 

[galactus]

Evaluate the double integral of (3x+4y^2) where R is the region in the upper half-plane bounded by the circles x^2 = y^2 = 1 and x^2 + y^2 = 4. How would I find 'R'? And where the heck does the extra r term come from?

Use \(\displaystyle x=rcos({\theta}), \;\ y=rsin({\theta})\)

The 'extra r' comes from the area of a sector of a circle formula. \(\displaystyle \frac{1}{2}r^{2}{\theta}\)

Instead of dividing the region up into strips, as in rectangular coordinates, we divide it up into 'wedges'.

If we choose \(\displaystyle (r_{k}, \;\ {\theta}_{k})\) as the center of the kth polar rectangle and this rectangle has a central angle of \(\displaystyle {\Delta}_{k}\) and a 'radial thickness' of \(\displaystyle {\Delta}r_{k}\), then the inner radius of the polar rectangle is \(\displaystyle r_{k}-\frac{1}{2}{\Delta}r_{k}\) and the outer radius is \(\displaystyle r_{k}+\frac{1}{2}{\Delta}r_{k}\).

Now, treat the area as the difference of the two sectors we get: \(\displaystyle \frac{1}{2}(r_{k}+\frac{1}{2}{\Delta}r_{k})^{2}{\Delta}{\theta}_{k}-\frac{1}{2}(r_{k}-\frac{1}{2}{\Delta}r_{k})^{2}{\Delta}{\theta}_{k}\)

Which simplifies down to: \(\displaystyle r_{k}{\Delta}r_{k}{\Delta}{\theta}_{k}\)

Therefore, whence, we have: \(\displaystyle \int\int{f({r, \;\ {\theta})}dA=\lim_{n\to\infty}\sum_{k=1}^{n}f(r_{k}, \;\ {\theta}_{k})r_{k} \;\ {\Delta}_{k} \;\ {\Delta}{\theta}_{k}\)

So, we have the familiar: \(\displaystyle \int_{\alpha}^{\beta}\int_{r_{1}({\theta})}^{r_{2}({\theta})}f(r,{\theta})rdrd{\theta}\)

Now, see why the 'extra r' is there?.

Your region is \(\displaystyle \int_{0}^{\pi}\int_{1}^{2}\left[3r^{2}cos{\theta}+4r^{3}sin^{2}{\theta}\right]drd{\theta}\)

Here's a haphazard diagram. Perhaps you can see where the 1/2 way point is and why we add and subtract to get our integration formula.