Homework Help: Double integral

1. May 11, 2010

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

Last edited: May 11, 2010
2. May 11, 2010

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?

3. May 11, 2010

lanedance

so what was the question?

4. May 11, 2010

phyzmatix

I think it might actually be

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

5. May 11, 2010

lanedance

yeah i think you're right

6. May 11, 2010

xstetsonx

my teacher said you have to separate the area into to parts in order to solve it and i don't understand why

sorry about my crappy drawing again

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Last edited: May 11, 2010
7. May 11, 2010

phyzmatix

What exactly was the problem statement?

8. May 11, 2010

Staff: Mentor

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.

9. May 11, 2010

xstetsonx

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

10. May 11, 2010

phyzmatix

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

11. May 11, 2010

xstetsonx

thx phyzmatrix anyways

12. May 11, 2010

Staff: Mentor

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.

13. May 11, 2010

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

14. May 11, 2010

Staff: Mentor

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?

15. May 11, 2010

xstetsonx

1? cos^2+sin^2????

16. May 11, 2010

Staff: Mentor

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

17. May 11, 2010

xstetsonx

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

18. May 11, 2010

Staff: Mentor

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

19. May 11, 2010

xstetsonx

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

20. May 11, 2010

Staff: Mentor

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.

21. May 11, 2010

xstetsonx

(2)^.5, 45 degrees?

22. May 11, 2010

Staff: Mentor

Yes. And what is 45 degrees in radians?

23. May 11, 2010

xstetsonx

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

24. May 11, 2010

Staff: Mentor

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?

25. May 11, 2010

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?