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Math Help - Finding the mass of a lamina given the density and the description of the lamina

  1. #1
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    Finding the mass of a lamina given the density and the description of the lamina

    Let W be the lamina in the plane enclosed by the curve xy = 1 and by the lines y = (1/3) x; y = 5x, i. e. ,
    W := [(x; y) | (1/3) x \le y \le 5x; x \ge 0; xy \le 1 ].
    Given that the density p(x; y) of the lamina is given by p(x; y) = x^2 + y^2 , calculate the mass of the
    lamina.
    the substitution u = xy; v = y/xcould be used or otherwise.
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  2. #2
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    45 posts and not a hint of your own work?!

    Here's a piece. You explain it to me and provide the other piece and the total evaluation.

    \int_{0}^{\frac{1}{\sqrt{5}}}\int_{\frac{x}{3}}^{5  \cdot x}\left(x^{2}+y^{2}\right)\;dydx
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    But that suggested substitution does make for an interesting simplification. The Jacobian gives dxdy= vdudv so the integrand becomes (u+ uv^2)dudv.
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    Quote Originally Posted by TKHunny View Post
    45 posts and not a hint of your own work?!

    Here's a piece. You explain it to me and provide the other piece and the total evaluation.

    \int_{0}^{\frac{1}{\sqrt{5}}}\int_{\frac{x}{3}}^{5  \cdot x}\left(x^{2}+y^{2}\right)\;dydx
    I think that integral you gave will give the density of the lamina, and we then have to find the surface area of it, then MASS= SURFACE AREA * DENSITY

    however I'm not sure how you came to this integral and chose those limits, and do I have to apply dy to the inside integral and then dx to the result of the inside?

    I drew the lamina and I think the upper and lower limit you chose for the inside uses the space between the straight lines and then on the outside integral I do not know how you chose 0 and 1/\sqrt{5} but I know that the line y=5x meet the curve at (\sqrt{5}/5 , \sqrt{5})
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    But that suggested substitution does make for an interesting simplification. The Jacobian gives dxdy= vdudv so the integrand becomes (u+ uv^2)dudv.
    once I make that substitution and the integrand becomes (u+ uv^2)dudv

    how do I solve the integral and also do I have to change the limits,

    I think I have to integrate the inside one first w.r.t du then the outer one w.r.t dv?
    limits of the inside will change using y=u/x and the outer one will use x=y/v
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