Finding the Equation of the Hypotenuse of a Triangle in Multi-Variable Calculus

[Larrytsai]

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

 

[Mark44]

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

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]

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

 

[Mark44]

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

That is NOT the equation of the hypotenuse of the triangle - it's the equation of a circle centered at the origin.

The hypotenuse of the triangle is a straight line segment. Do you know how to find the equation of a line?