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Double integral question?
Solve the double integral I=SSd (xy) dA where d is a triangle with vertices (0,0)(5,0)(0,3)
S is supposed to be an integral sign lol...
I keep getting 28.125 but thats wrong...
1 Answer
- kbLv 71 decade agoFavorite Answer
The triangular region is bounded between the the x,y axes and y = -3x/5 + 3.
Thus, ∫∫ xy dA
= ∫(x = 0 to 5) ∫(y = 0 to 3 - 3x/5) xy dy dx
= ∫(x = 0 to 5) xy^2/2 {for y = 0 to 3 - 3x/5} dx
= (1/2) ∫(x = 0 to 5) x(3 - 3x/5)^2 dx
= (1/2) ∫(x = 0 to 5) (3/5)^2 * x(5 - x)^2 dx
= (9/50) ∫(x = 0 to 5) (x^3 - 10x^2 + 25x) dx
= (9/50) (x^4/4 - 10x^3/3 + 25x^2/2) {for x = 0 to 5}
= (9/50) (625/12)
= 75/8.
I hope this helps!
