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The density at each point of a 1 cm square plate is $16+r^4$ g/cm2, where $r$ is the distance in cm from the point to the centre of the plate. What is the mass of the plate?

I know that mass is the double integration of the density function. My problem is that the function of the density is in polar coordinates but the plate is a square. Should I convert the density function into Cartesians? How do I put the limits after the conversion?

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  • $\begingroup$ actually, what you do is do the integral from $\theta = 0$ to $\theta = \pi/4,$ then multiply by 8. It is not so difficult to write the line $x = 1/2$ in polar coordinates. $\endgroup$ – Will Jagy Sep 18 '17 at 2:40
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Just use Cartesian coordinates supposing the square is centered at $0$. You will obtain the integral $$\int_{-1/2}^{1/2}\int_{-1/2}^{1/2} (16+(x^2+y^2)^2)dxdy,$$ which is easy to calculate.

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  • $\begingroup$ Just observe I am using $r = (x^2 + y^2)^{1/2}$ and $dMass = (16 + r^4)dxdy$. $\endgroup$ – Hugocito Sep 18 '17 at 2:48
  • $\begingroup$ Is it the same when I do the $4\int_0^\frac{1}{2}\int_0^\frac{1}{2}16+(x^2+y^2)^2dxdy$ $\endgroup$ – Katharine Kim Sep 18 '17 at 2:49
  • $\begingroup$ Yes. The reason is very simple, the function you are integrating is even with respect to $x$ and $y$. $\endgroup$ – Hugocito Sep 18 '17 at 2:52

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