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Thread: 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 $\displaystyle W$ be the lamina in the plane enclosed by the curve $\displaystyle xy = 1$ and by the lines $\displaystyle y = (1/3) x$;$\displaystyle y = 5x$, i. e. ,
    $\displaystyle W := [(x; y) |$ $\displaystyle (1/3) x$$\displaystyle \le$ y $\displaystyle \le$$\displaystyle 5x$; $\displaystyle x$ $\displaystyle \ge$ $\displaystyle 0$; $\displaystyle xy \le 1$ ].
    Given that the density p(x; y) of the lamina is given by $\displaystyle p(x; y) = x^2 + y^2$ , calculate the mass of the
    lamina.
    the substitution u$\displaystyle = xy; v = y/x$could be used or otherwise.
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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.

    $\displaystyle \int_{0}^{\frac{1}{\sqrt{5}}}\int_{\frac{x}{3}}^{5 \cdot x}\left(x^{2}+y^{2}\right)\;dydx$
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  3. #3
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    But that suggested substitution does make for an interesting simplification. The Jacobian gives $\displaystyle dxdy= vdudv$ so the integrand becomes $\displaystyle (u+ uv^2)dudv$.
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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.

    $\displaystyle \int_{0}^{\frac{1}{\sqrt{5}}}\int_{\frac{x}{3}}^{5 \cdot x}\left(x^{2}+y^{2}\right)\;dydx$
    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 $\displaystyle MASS= SURFACE AREA * DENSITY$

    however I'm not sure how you came to this integral and chose those limits, and do I have to apply $\displaystyle dy$ to the inside integral and then $\displaystyle dx$ 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 $\displaystyle 0 and 1/\sqrt{5}$ but I know that the line $\displaystyle y=5x$ meet the curve at $\displaystyle (\sqrt{5}/5 , \sqrt{5})$
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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 $\displaystyle dxdy= vdudv$ so the integrand becomes $\displaystyle (u+ uv^2)dudv$.
    once I make that substitution and the integrand becomes $\displaystyle (u+ uv^2)dudv$

    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 $\displaystyle y=u/x$ and the outer one will use $\displaystyle x=y/v$
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