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The triangle has the following vertices:

$$(0,0),(c,0),(0,c)$$

I drew this figure and found the slope and the line:

$$m=\frac{0-c}{c-0}\frac{-c}{c}=-1$$

And was then able to form the following line:

$$y-0=-1(x-c)$$

$$y=-x+c$$

Since the mass is calculated with the following formula:

$$M=\iint_D \rho(x,y)\;dA = \int^{c}_{0}\int^{-x+c}_{0}x^2+y^2\;dy\;dx$$

I then converted this to cylindrical coordinates for ease of computation:

$$r\sin\theta=-r\cos\theta+c$$

$$r=\sqrt{c}$$

My integral thus becomes:

$$M=\int^{\frac{\pi}{2}}_{0}\int^{\sqrt{c}}_{0}r^2\;r\;dr\;d\theta$$

Is this correct? I'm mainly lost at what the limits of integration should be for my outer integral.

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2 Answers 2

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Try not converting to cylindrical coordinates! Your integrand is just a polynomial ($x^2+y^2$) which is really easy to integrate so doesn't need to be simplified. The surface however is a triangle, which does not lend well to cylindrical coordinates. You should be able to calculate the integral easily directly from your Cartesian coordinates ($x$ and $y$)!

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  • $\begingroup$ Right, but the integral becomes more difficult (or I became more lazy) once I started to evaluate the outer integral with respects to $x$ b $\endgroup$
    – nullByteMe
    Dec 12, 2015 at 21:14
  • $\begingroup$ I guess I should have factor out the fraction and integrated the second term separately. Hmm $\endgroup$
    – nullByteMe
    Dec 12, 2015 at 21:18
  • $\begingroup$ It's hard to know exactly what you're getting at since I cant see your calculations :). But after you integrate the inner integral you will still have a polynomial. Substitute for the limits, and you will have a polynomial once again (of degree 3.) You might want to make sure that you expand everything and simplify, before integrating the second time. $\endgroup$ Dec 12, 2015 at 21:22
  • $\begingroup$ So the integral is correctly expressed prior to my conversions to cylindrical coordinates? $\endgroup$
    – nullByteMe
    Dec 12, 2015 at 21:24
  • $\begingroup$ Yes it is. Also remember that x is a constant when you integrate w.r.t. dy, and vice versa. $\endgroup$ Dec 12, 2015 at 21:26
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HINT: For triangle bounded by x-, y- axes and line $$ x + y =c, \; Area =\int_0 ^c y\, dx .. $$

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