Double integral help!
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# Double integral help!

[From: ] [author: ] [Date: 11-10-07] [Hit: ]
= -1/48.I hope this helps!......
evaluate ∫∫R xy if R is the region bounded by y = x^3 and y = x^2.

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I am assuming you want to evaluate ∫∫R xy dA.

y = x^3 and y = x^2 intersect when:
x^3 = x^2 ==> x = 0 and x = 1.

Note that R is a type-I region because we can represent R as:
R = {(x, y) | 0 <= x <= 1, x^2 <= y <= x^3}.

So, we have:
∫∫ R xy dA = ∫(from x=0 to 1) ∫(from y=x^2 to x^3) xy dy dx
= ∫(from x=0 to 1) [(1/2)xy^2] (eval from y=x^2 to x^3) dx, by doing the inner integral
= 1/2 ∫(from x=0 to 1) x[(x^3)^2 - (x^2)^2] dx
= 1/2 ∫(from x=0 to 1) (x^7 - x^5) dx, by expanding
= (1/2)[(1/8)x^8 - (1/6)x^6] (eval from x=0 to 1), by evaluating this integral
= (1/2)[(1/8 - 1/6) - (0 - 0)]
= -1/48.

I hope this helps!
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