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Double integral

  • #1

Homework Statement




Evaluate the integral

∫∫R y√(x^2 + y^2) dA

with R the region {(x, y) : 1 ≤ x^2 + y^2 ≤ 2, 0 ≤ y ≤ x.}

Homework Equations





The Attempt at a Solution



Solution:

∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx

∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0) dx

1/3 ∫(2 to 1) [(x^2 + y^2)^(3/2) - (x^2)^(3/2)] dx

1/3 x 2/5 [(x^2 + y^2)^(5/2) - (x^2)^(5/2)] |(x=2 to x=1)

final answer: 4(124√2 - 33)/15

Is this the correct answer??
 

Answers and Replies

  • #2
Dick
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No, I don't think that's right. For one thing the number is far too large to fit in the annulus 1<=r<=2. And for another, why don't you use the polar coordinates the problem is so obviously screaming for?
 
  • #3
so

∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx
∫(2 to 1)∫(x to 0) ((r sin(theta)(r^2)^1/2)r dr dtheta
∫(2 to 1)∫(x to 0) ((r^2 sin(theta)(r^3)^1/2) dr dtheta

Like this?
 
  • #4
Dick
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Sort of like that, but without the errors. Where did (r^3)^(1/2) come from? And you are integrating dtheta. What are the theta limits?
 
  • #5
I got the (r^3)^(1/2) from multiplying the r on the outside to the inside.. the (1/2) is the sqrt and the r^3 is from multiplying it with the r..

also the theta limit would be r to 0, rather than x to 0??
 
  • #6
Dick
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I got the (r^3)^(1/2) from multiplying the r on the outside to the inside.. the (1/2) is the sqrt and the r^3 is from multiplying it with the r..

also the theta limit would be r to 0, rather than x to 0??
The r power is wrong. You started with (r^2)^(1/2). Not (r)^(1/2). Why do you think the theta limits are 0 to r? Is that just a random guess? Did you draw a picture of the region?
 
  • #7
oh its a circle that would go from 2pie to 0.. but iam still confused about the other part
 
  • #8
Dick
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oh its a circle that would go from 2pie to 0.. but iam still confused about the other part
????? What do you think it looks like??? 0<=y<=x. Start picking some values of x and sketching the region.
 
  • #9
What if i tried not to use polar and did this..

from above we have:

∫(2 to 1)∫(x to 0) y√(x^2 + y^2) dy dx

∫(2 to 1) 1/3[(x^2 + y^2)^(3/2)] |(y=x and y=0) dx

1/3 ∫(2 to 1) x^3 dx

1/3 x 1/4 x^4 | (x = 2 and x =1)

final answer: 5/4
 
  • #10
Dick
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Just because you aren't using polar coordinates doesn't mean you can skip understanding the problem. Sketch the region!!!! If you do this you will realize you can't do it dx*dy using a single set of limits. You'd have to split the region up. Polar is easier.
 
  • #11
iam not so sure i know how to sketch it.
 
  • #12
Defennder
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It's kind of tedious if you don't use polar coordinates. If you do by polar, you won't have to worry about finding the limits of y as a function of x. The limits in polar coordinates are just numbers.
 
  • #13
Dick
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iam not so sure i know how to sketch it.
Start with 1<=r<=2. Any ideas what that might look like? Then try 0<=y<=x. As I said, pick a value of x, say x=1. What values of y work? Draw them. Try x=2. Draw them. Etc. Until you get the general picture.
 
  • #14
Defennder
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iam not so sure i know how to sketch it.
Well, you have x^2 + y^2 ≤ 2. Consider the case x^2 + y^2 = 2. This represents a circle of radius sqrt(2). Consider also x^2 + y^2 = 1. This is circle of radius 1. The region you want is in between these 2 in the inequality. How does this translate graphically? For y<=x, draw the line y=x and determine the locus of points where y<x. The intersection of these 2 regions is the region over which the double integration is performed.
 
  • #15
hm what I came up with, which is probably wrong, is almost like a triangle in the first quad...

so therefore you would go from (pie/2)/2 = pie/4 to 0??
 
  • #16
Defennder
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Yes, in the first quadrant it is a triangular region extending to infinity. I'm assuming you haven't found the other region corresponding to 1 ≤ x^2 + y^2 ≤ 2. But what about the other quadrants?
 
  • #17
Dick
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hm what I came up with, which is probably wrong, is almost like a triangle in the first quad...

so therefore you would go from (pie/2)/2 = pie/4 to 0??
Yessss! The region is 1<=r<=2 and 0<=theta<=pi/4. It is a wedge shaped triangular region intersected with a ring shaped region.
 
  • #18
Dick
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hm what I came up with, which is probably wrong, is almost like a triangle in the first quad...

so therefore you would go from (pie/2)/2 = pie/4 to 0??
Yessss! The region is 1<=r<=2 and 0<=theta<=pi/4. It is a wedge shaped triangular region. See how helpful sketching the region is?
 

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