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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.