Double integrals question
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Double integrals question

[From: ] [author: ] [Date: 12-10-19] [Hit: ]
-∫∫R xy√(x^2 + y^2) dA on [0,1] x [0,= ∫(0 to 1) x/3*(x^2 + y^2)^(3/2) eval.= 1/15 * (x^2 + 1)^(5/2) - (x^5)/15 eval.= 1/15 * (2)^(5/2) - 1/15 - 1/15 = 0.38 - 2/15 = 0.......
How would I integrate this:

∬R xy√(x^2 + y^2) R = [0,1]x[0,1]

I really have no idea.

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∫∫R xy√(x^2 + y^2) dA on [0,1] x [0,1]

= ∫(0 to 1) ∫(0 to 1) xy√(x^2 + y^2) dy dx

= ∫(0 to 1) x/3*(x^2 + y^2)^(3/2) eval. from y = 0 to y = 1 dx

= ∫(0 to 1) x/3 * (x^2 + 1)^(3/2) - (x^4)/3 dx

= 1/15 * (x^2 + 1)^(5/2) - (x^5)/15 eval. from 0 to 1

= 1/15 * (2)^(5/2) - 1/15 - 1/15 = 0.38 - 2/15 = 0.24379
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