Double Integrals: Solving f(x,y)= x2 + y2 in a Triangle Region

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SUMMARY

The discussion focuses on solving the double integral of the function f(x,y) = x² + y² over a triangular region defined by the vertices (0,0), (1,0), and (0,1). The initial attempt at the solution involved integrating with respect to y first, leading to confusion regarding the limits of integration. The correct approach requires careful consideration of the integration order and the boundaries defined by the line y = 1 - x. The final answer, as confirmed by the textbook, is 1/6.

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Homework Statement



f(x,y) = x2 + y2 over the triangular region with vertices (0,0), (1,0), (0,1)

Homework Equations





The Attempt at a Solution



\intfrom 0 to 1-x\intfrom 0 to 1 of x2 + y2 dydx = \intfrom 0 to 1-x of (1/3)y3 evaluated from 0 to 1 dx
= \int from 0 to 1-x of (1/3) dx
= (1/3)x evaluated from 0 to 1-x
= (1/3)(1-x)

According to the book, the answer is 1/6... How do I know what the value of x is?
 
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Is there another way you can define the line 1-x? Maybe in terms of y :wink:.

Oh, and if you choose to do so you might need to change the order in which you're integrating :wink:.
 
You've written the equivalent of this:

\int^{1-x}_0 f(x) dx.

You can't "let x go from 0 to (1-x)." That doesn't make any sense.

Try writing down your double integral again, but pay more careful attention to which direction you're integrating and what order your dy and dx are in.
 

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