Limits after mapping in double integral

[nickthegreek]

Homework Statement

I have the double integral,

∫∫sqrt(x^2+y^2) dxdy, and the area D:((x,y);(x^2+y^2)≤ x)

Homework Equations

The Attempt at a Solution

By completing the squares in D we get that D is a circle with origo at (1/2,0), and radius 1/2. Then I tried changing the variables to x=r cosθ+1/2, y=r sinθ and J(r,θ)=r which leads to a not so nice integrand, and stop.

By changing to polar coordinates directly we get that D transforms into E:((r,θ);r^2≤ r cosθ)) which obv equals r≤cosθ, and the integrand r^3, which is nice. Now to my question. What do I know of "E"? What would it look like? What´s the limits?

 

[Vargo]

r varies from 0 to cos(theta). This disk is contained in the right half plane, so x must be nonnegative. That tells you what the restrictions on theta should be. So you get an iterated integral.

 

[nickthegreek]

Hi Vargo, thanks for answering!

I understand that this means that θ varies from -π/2 to π/2, which is the correct answer. Tho I don't understand why it does. I probably don't understand the mapping fully, I think of E as some rectangle in some r,θ-plane, and I don't understand why we look at x after we´ve mapped it into r,θ? Could you explain it a bit more detailed? Really appreciate any help I get...

 

[Vargo]

Well, first it is not really a rectangle in the r theta plane. The upper limit of r is not constant but rather depends on theta.

Second, you don't really want to look too much in the "r theta plane". It is better to just have the one picture in the euclidean plane and just think of r theta as alternative coordinates for this plane (at least it makes more sense for me this way). The circle is in the right half plane so yes theta varies from -pi/2 to pi/2 (because those are the angles of the points in the right half plane). Alternatively, you could consider theta to vary from 0 to pi/2 and then 3pi/2 to 2pi, but it is more convenient to let it vary over the connected interval -pi/2 to pi/2.

Think of a radial vector pointing in the direction theta whose length is cos(theta). As theta varies from -pi/2 to pi/2, this radial vector sweeps out the interior of the disk. So in your integral, you first integrate with respect to r (0 to cos(theta)) and then with respect to theta. And of course you integrate with respect to r dr dtheta.

 

[nickthegreek]

Thx! I need to get the picture of this straight in my head, after I've sorted the picture out I have no problems calculating it. Is it correct if I think of it this way?

At first we have the circle, let's call it D, at (1/2,0) with r=1/2. We make the substitution and move into (0,0). We have a radius, which we know is less than or equal to cos theta. To find out what theta is we look for the angle in which we need to sweep to cover D, which obv is
-pi/2 to pi/2. r would work like a radar and beep when he's over our D, and he would only stretch out to the length needed to cover it (how long is none of our business). Would thinking of it this way cause me trouble? :)

 

[Vargo]

Sounds about right.