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The density at each point of a 1cm square plate is $1+r^6$ grams per square cm, where r is the distance in cm from the point to the certer of the plate. What is the mass of the plate?

So far what I have done as below, but it looks crazy: $$mass=4 \int_0^{\frac{1}{2}}\int_0^{\frac{1}{2}}{1+(\sqrt{x^2+y^2})^6} dx\,dy = \frac{283}{280}$$

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  • $\begingroup$ @YvesDaoust edited $\endgroup$ Feb 4, 2018 at 11:47
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    $\begingroup$ That's correct. Expand the sixth power and you'll get a bivariate polynomial, not so terrible. $\endgroup$
    – user65203
    Feb 4, 2018 at 11:55
  • $\begingroup$ Is there a better/ smarter way yo solve the question? Polar coordinate doesn't seem to work well. $\endgroup$ Feb 4, 2018 at 12:03
  • $\begingroup$ if intergrals is correct then mass :${\frac{283}{280}}$ $\endgroup$ Feb 4, 2018 at 12:18
  • $\begingroup$ @KatharineKim: what's wrong with the Cartesian integral ? $\endgroup$
    – user65203
    Feb 4, 2018 at 12:52

2 Answers 2

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I guess the plate is the square with vertices $(\pm 1/2,\pm 1/2)$(?)

Instead of cutting it in $1/4$, cut it in $1/8$:

$$\mbox{mass } = 8 \int_0^{\pi/4} \int_0^{\sec\theta/2} (1+r^6) r\; dr \; d\theta.$$

Even powers of $\sec \theta$ are easy to integrate.

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Let $r=\sqrt{x^2+y^2}$, Notice that $1+(\sqrt{x^2+y^2})^6$ is a odd function. The mass is 4 times the 1st quadrant.

$$mass= 4 \int_0^{\frac{1}{2}}\int_0^{\frac{1}{2}}{1+(\sqrt{x^2+y^2})^6} dx\,dy$$

Expend the square root and do integration on the polynomial.

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