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Evaluate this double integral:

$\int \int _{R} x dx dy$ where $R$ is the area of a quadrilateral of vertex $(0,-1),(5,-1),(3,1),(2,1).$

When i do the double integral i get the answer $15$. In the solution the answer is also $15$. What i don't understand is why when i calculate the area of this quadrilateral using basic geometry, I get an area of $6$. Am I doing something wrong here? Is it normal to get different answer?

Thank you.

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2 Answers 2

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The reason is that the integral you evaluated doesn't give the area of the quadrilateral $R$. Indeed, called $A$ its area, you have $$\int\int_R dx dy=A$$ and $$\int\int_R x dx dy\neq \int\int_R dx dy.$$

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  • $\begingroup$ The region isn’t a square. It’s a quadrilateral. $\endgroup$
    – amd
    Nov 24, 2015 at 19:01
  • $\begingroup$ what are we evaluating then? $\endgroup$
    – spexel
    Nov 24, 2015 at 19:01
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    $\begingroup$ @spexel You could interpret the integral as computing the mass of a quadrilateral lamina that has increasing density to the right. $\endgroup$
    – amd
    Nov 24, 2015 at 19:03
  • $\begingroup$ Moment of inertia around the $y$-axis (of a uniformly dense quadrilateral)? Oops, that'd be the second moment of mass; this is the first moment. I'm not sure what the physical interpretation of this quantity would be. $\endgroup$
    – Brian Tung
    Nov 24, 2015 at 19:07
  • $\begingroup$ @BrianTung see TacoTesseract’s answer for another physical interpretation of the integral. $\endgroup$
    – amd
    Nov 24, 2015 at 20:14
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What your integral finds is the volume under the surface $z = x$ above your quadrilateral. If you wanted to find the area of your domain (the quadrilateral) use $z = 1$ as your integrand.

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