Double integral

helpm3pl3ase · May 18, 2008

1. The problem statement, all variables and given/known data

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.}

2. Relevant equations

3. 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??

Dick · May 18, 2008

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?

helpm3pl3ase · May 19, 2008

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?

Dick · May 19, 2008

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?

helpm3pl3ase · May 19, 2008

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??

Dick · May 19, 2008

helpm3pl3ase said: ↑

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?

helpm3pl3ase · May 19, 2008

oh its a circle that would go from 2pie to 0.. but iam still confused about the other part

Dick · May 19, 2008

helpm3pl3ase said: ↑

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.

helpm3pl3ase · May 19, 2008

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

Dick · May 19, 2008

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.

helpm3pl3ase · May 19, 2008

iam not so sure i know how to sketch it.

Defennder · May 19, 2008

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.

Dick · May 19, 2008

helpm3pl3ase said: ↑

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.

Defennder · May 19, 2008

helpm3pl3ase said: ↑

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.

helpm3pl3ase · May 19, 2008

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??

Defennder · May 19, 2008

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?

Dick · May 19, 2008

helpm3pl3ase said: ↑

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.

Dick · May 19, 2008

helpm3pl3ase said: ↑

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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Double integral

helpm3pl3ase

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Dick

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Dick

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Dick

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Defennder

Dick

Defennder

helpm3pl3ase

Defennder

Dick

Dick

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