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Exercise 521Calculus › Multivariable calculus › Multiple integralsHigh schoolAdded 3 Nov 2026

A symmetric triangle integral

#symmetry#triangle#double integral

Statement

Compute I=∬D(x+y) dx dy\displaystyle I=\iint_D (x+y)\,dx\,dy where DD is the triangle with vertices (0,0),(2,0),(0,2)(0,0),(2,0),(0,2).

Hint

The triangle is symmetric in xx and yy: compute ∬Dx\iint_D x only and multiply by 22.

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Solution

Setup. D={(x,y):0≤x≤2, 0≤y≤2−x}D=\{(x,y):0\le x\le2,\ 0\le y\le2-x\}.

Compute ∬Dx\iint_D x.
∫02∫02−xx dy dx=∫02x(2−x) dx=[x2−x33]02=4−83=43.\int_0^2\int_0^{2-x}x\,dy\,dx=\int_0^2 x(2-x)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}.

Symmetry. ∬Dy=43\iint_D y=\frac43, so I=83I=\frac83.

Conclusion. I=8/3I=8/3.