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Math Help - Double Integration and its Applications

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    Double Integration and its Applications

    1. \iint_{R}e^{x+y}~dA where R: |x| + |y| <= 2

    2. A lamina occupies the part of the disk x^2 + y^2 <= 1 in the first quadrant. Find its center of mas if the density at any given point is proportional to the square of its distance from the origin.
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    Quote Originally Posted by algebrapro18 View Post
    1. \iint_{R}e^{x+y}~dA where R: |x| + |y| <= 2

    [snip]
    The region of integration in the xy-plane is bounded by the following four lines:

    x + y = 2 when x > 0 and y > 0, that is, the line y = -x + 2 in the first quadrant.

    -x + y = 2 when x < 0 and y > 0, that is, the line y = x + 2 in the second quadrant.

    -x - y = 2 when x < 0 and y < 0, that is, the line y = -x - 2 in the third quadrant.

    x - y = 2 when x > 0 and y < 0, that is, the line y = x - 2 in the fourth quadrant.

    The region is therefore a diamond centered on the origin.

    Life is much easier if you now make the following coordinate transformation (suggested by this region):

    u = x + y .... (1)

    v = x - y .... (2)

    It follows from (1) and (2) that x = \frac{u+v}{2} and  y = \frac{u-v}{2}. Therefore the Jacobian of the transformation is ........

    The region of integration in the uv-plane is a square. The integrations are trivial.
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    Quote Originally Posted by algebrapro18 View Post
    [snip]
    2. A lamina occupies the part of the disk x^2 + y^2 <= 1 in the first quadrant. Find its center of mas if the density at any given point is proportional to the square of its distance from the origin.
    Where are you stuck here? Show the working you've done.
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    Quote Originally Posted by mr fantastic View Post
    Where are you stuck here? Show the working you've done.
    I haven't done anything I don't know how to start.

    I know the center of mass:

    x = \iint_{R}xp(x,y)~dA / mass of the Lamina
    y = \iint_{R}yp(x,y)~dA / mass of the Lamina

    but I don't know how to find P(x,y) and the mass of the Lamina.
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    Quote Originally Posted by algebrapro18 View Post
    I haven't done anything I don't know how to start.

    I know the center of mass:

    x = \iint_{R}xp(x,y)~dA / mass of the Lamina
    y = \iint_{R}yp(x,y)~dA / mass of the Lamina

    but I don't know how to find P(x,y) and the mass of the Lamina.
    You're given \rho (x,y) !

    \rho (x,y) = k(x^2 + y^2) where k is the constant of proportionality.

    And you should know that

    M = \int \int_{R_{xy}} \rho (x,y) \, dx \, dy

    where R_{xy} is the region in the xy-plane the lamina is spread over ......

    And R_{xy} is also given to you. It's the part of the disk x^2 + y^2 \leq 1 in the first quadrant.
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    I really don't feel like tying out all my work but can you check my answers?

    M = (pi*k)/8
    Center of mass = 8/(5*pi)
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    Quote Originally Posted by algebrapro18 View Post
    I really don't feel like tying out all my work but can you check my answers?

    M = (pi*k)/8
    Center of mass = 8/(5*pi)
    If you mean that x-coordinate = y-coordinate = 8/(5*pi) then your calculations are correct.
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