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∬_(D)(2x+y)dxdy,D:x+y=3;y=0;x=0

∬D(2x+y)dxdy,D:x+y=3;y=0;x=0 \iint_{D}(2 x+y) d x d y, D: x+y=3 ; y=0 ; x=0

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Q. ∬D(2x+y)dxdy,D:x+y=3;y=0;x=0 \iint_{D}(2 x+y) d x d y, D: x+y=3 ; y=0 ; x=0
  1. Determine region D: Determine the region D for integration.\newlineThe region D is bounded by the lines x+y=3x+y=3, y=0y=0, and x=0x=0. This forms a right triangle in the first quadrant with vertices at (0,0)(0,0), (3,0)(3,0), and (0,3)(0,3).
  2. Set up double integral: Set up the double integral. Since yy ranges from 00 to 33 (from the line y=0y=0 to the line x+y=3x+y=3 where x=0x=0), and xx ranges from 00 to 3−y3-y (from x=0x=0 to x+y=3x+y=3), the integral is set up as: 0011.
  3. Integrate with respect to x: Integrate with respect to x. ∫x=03−y(2x+y) dx=[x2+yx]x=03−y\int_{x=0}^{3-y} (2x+y) \, dx = [x^2 + yx]_{x=0}^{3-y} = [(3−y)2+y(3−y)]−[0+0][(3-y)^2 + y(3-y)] - [0 + 0] = (3−y)2+3y−y2(3-y)^2 + 3y - y^2.
  4. Simplify expression: Simplify the expression.\newline(3−y)2+3y−y2=9−6y+y2+3y−y2(3-y)^2 + 3y - y^2 = 9 - 6y + y^2 + 3y - y^2\newline=9−3y.= 9 - 3y.

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