The discussion revolves around evaluating a double integral over a triangular region defined by specific vertices in a multi-variable calculus context. Participants are focused on determining the correct bounds for the integral based on the geometry of the triangle.
The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the nature of the hypotenuse, indicating that the current understanding may need to be reassessed.
There is a noted misunderstanding regarding the geometry of the triangle, particularly in distinguishing between the equation of a line segment and that of a circle. This may affect the determination of the correct bounds for the integral.
No they wouldn't. If the region were a rectangle, those would be the bounds, but your region is a triangle. Think about the equation of the hypotenuse of the triangle.Larrytsai said:Homework Statement
Evaluate the double integral I=int(int(D)( xydA) where D is the triangular region with vertices (0,0)(5,0)(0,3).
The Attempt at a Solution
I was wondering if my bounds for x would be 0 to 5 and y to be 0 to 3
That is NOT the equation of the hypotenuse of the triangle - it's the equation of a circle centered at the origin.Larrytsai said:k so my hypotnuse is x^2+y^2 = sqrt(34), then i isolate for x and have x=sqrt(sqrt(34)-y^2) so my bounds for x would be 0 to x=sqrt(sqrt(34)-y^2) , and y would be 0 to 3