Question

Evaluate the double integral int int y^3 dA where D is the triangular region with vertices (0,1), (7,0) and (1,1).

Answer 1

Answer:

1/5

Step-by-step explanation:

Refer to the image for the region. I've chosen to "scan" the region horizontally for convenience sake, so x varies for values between the green and the red line, and y varies from 0 to 1.

The two lines have equations $x=7-6y$ (red line) and $x=7-7y$ (green line). Since we're integrating first with respect to x, it's easier to write the equation like that and not as $y=mx+q$ since they will become our new limits of integration

Our double integral will become

$\int \limits_0^1 \ \ \int \limits_{7-7y}^{7-6y}y^3dxdy = \int \limits_0^1 y^3x|\limits_{7-7y}^{7-6y} dy =\\\int \limits_0^1 y^3[(7-6y)-(7-7y)]dy = \int \limits_0^1y^3(y)dy =\\\int \limits_0^1 y^4dy =\frac15y^5|\limits_0^1=\frac15$