Surface integral or Divergence Theorem confused?

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SUMMARY

The discussion centers on evaluating the double integral ∫∫ xy dA over the region R, bounded by the line y=x-1 and the parabola y²=2x+6. The correct limits of integration were determined to be from y=-2 to y=4, with x ranging from y+1 to (y²-6)/2, resulting in a final value of 52/3. A key point of contention is the terminology used in the problem; the term "volume" is misleading as the integral represents an area calculation, not a volume. The suggestion to use the divergence theorem was deemed unnecessary for this specific integral.

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Homework Statement



Find the Volume
∫∫ xy DA
where R is the region bounded by by the line y=x-1 and the parabola y^2=2x+6.

Homework Equations



∫∫ xy dx dy

The Attempt at a Solution



first i found the intersection of the above equations . which is (5,4) to (-1,-2) . then i simple put the values in the limits of the integral
i-e

∫(from y= -2 to y=4) ∫(from x= y+1 to x= (y^2-6)/2 ) xy dx dy

and solve it and finaly got 52/3.

is this the right method and limits are correct or not ?
or i have to use here divergence theorem

can somebody explain the word VOLUME in the question ?
 
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The word "volume" is an error. Clearly that is a two dimensional figure so it has area not volume.
 
so that means my method is correct?

but someone says

If the problem really is to find a "volume", it should read ∫ ∫ |xy| dA because the integrand is not positive everywhere on the region given.

Can't be sure, but that's probably not what the author intended. He just wants you to integrate xy over the region. It has the form ∫ ∫ z(x, y) dA which is a volume integral if z(x, y) is non-negative over the region.
 

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