Understanding Polar Coordinate Integration

[xstetsonx]

can someone explain to me why my teacher divided the area into two?
I=$$\int$$1,0$$\int$$(2x-x^2)^.5,0 1/(x^2+y^2)^.5dydx
ugggggh i tried to use the latex...
anyway...

he used the polar coordinate to do this. once he turned it into polar coordinate, he divided the area into 2 bounded by (pi/4 - o d$$\theta$$) (1/cos$$\theta$$ - 0 for dr) and then + (d$$\theta$$ bounded by pi/2 - pi/4) (2cos bounded by 2 - 0)sorry i don't know how to make this more clear...ask me if you need to clarify

 

[lanedance]

see the correct (hopefully) tex below, click on it to see script, note you can put the whole equation in one set of tags
$$\int_0^1 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx$$

is this what you meant?

 

[lanedance]

so what was the question?

 

[phyzmatix]

see the correct (hopefully) tex below, click on it to see script, note you can put the whole equation in one set of tags
$$\int_0^1 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx$$

is this what you meant?

I think it might actually be

$$\int_0^1 \int_0^{\sqrt{2x-x^2}} \frac{1}{\sqrt{x^2+y^2}} dy dx$$

 

[lanedance]

yeah i think you're right

 

[xstetsonx]

my teacher said you have to separate the area into to parts in order to solve it and i don't understand whysorry about my crappy drawing again

 

[phyzmatix]

What exactly was the problem statement?

 

[Mark44]

The region over which integration takes place is the upper left quarter of a circle with radius 1 and centered at (1, 0). When you switch to a polar integral, the description for r changes, and this is why you need to write the polar form of your given integral with two integrals. For one integral, r ranges from 0 to the appropriate value on the line x = 1. For the other integral, r ranges from 0 to the appropriate value on the circle.

 

[xstetsonx]

then how you suppose to figure out the boundaries? and can you expand on the description for r changes??

 

[phyzmatix]

Enters Mark! You're in much better hands than mine now xstetsonx so I'm definitely stepping back here. Have fun!

 

[xstetsonx]

thx phyzmatrix anyways

 

[Mark44]

then how you suppose to figure out the boundaries? and can you expand on the description for r changes??

Have you drawn a sketch of the region? I have described the boundaries. All you need to do is change the Cartesian equation that represents each boundary to its polar form.

 

[xstetsonx]

yup under the attachment above. well you said appropriate i am not clear. my teacher he
said for d$$\theta$$ is from 0 to pi/4 for the first one...etc i don't understand how you get this value according to that graph

 

[Mark44]

The graph in your attachment is not very useful, as the circle part doesn't look much like part of a circle, and almost nothing is identified with its equation. To answer your question, what are the coordinates of the point where the vertical line intersects the circle? For the same point, what are the polar coordinates?

 

[xstetsonx]

To answer your question, what are the coordinates of the point where the vertical line intersects the circle? For the same point, what are the polar coordinates?

1? cos^2+sin^2?

 

[Mark44]

1? cos^2+sin^2?

Is that supposed to be a point? I'm looking for two numbers.

 

[xstetsonx]

i mean (1,1) that is where they intersect right?

 

[Mark44]

Right. Now, what are the polar coordinates of that point?

 

[xstetsonx]

em (cosx)^2+(sinx)^2?

 

[Mark44]

No. Polar coordinates are pairs of numbers (r, theta), with the first being the distance from the pole (origin) and the second the angle theta.

In any case, cos²(x) +sin²(x) is identically equal to 1.

 

[xstetsonx]

(2)^.5, 45 degrees?

 

[Mark44]

Yes. And what is 45 degrees in radians?

 

[xstetsonx]

MARK you are once again the awesomest guy on here
thanks alot

 

[Mark44]

Well, thanks!

So now you understand why the limits on one integral are 0 to pi/4 and for the other they are pi/4 to pi/2, right?

 

[xstetsonx]

yup yup everything makes lot senses now but i just want to know how he get the equation 1/cosx and 2cosx

if i am doing from scratch how do i know it is that equation?

 

[Mark44]

Equations have = in them, and I don't see any.

The line x = 1 (one boundary of the region of integration) can be converted to polar this way:
r*cos(theta) = 1 ==> r = 1/cos(theta) = sec(theta).

The circle's equation is x^2 + y^2 = 2x. Converting to polar, you get this:
r^2 = 2rcos(theta) ==> r = 2cos(theta), which I get by dividing both sides of the first equation by r.

 

[phyzmatix]

MARK you are once again the awesomest guy on here
thanks alot

Told you he's a legend! His input to homework threads (not necessarily my own) has helped me on more than one occasion.

Anyway, I'm waaaay off topic here!

 

[xstetsonx]

yea thanks to mark i got an A on my exam!

 

[Mark44]

That's great!