Question

Evaluate the double integral ∬∬Df(x,y)dA over the region D, where =sin⁡f(x,y)=sin(y) and D is the triangular region with vertices (0, 0), (0, 7), and (7, 0).

A) 49/2
B) 7/2
C) 21/2
D) 14/2

Answer 1

To evaluate the double integral over the triangular region D, set up the limits of integration and integrand, integrate with respect to y first, then integrate with respect to x.

Step-by-step explanation:

To evaluate the double integral ∫∫Df(x,y)dA over the region D, we need to find the limits of integration and the integrand.

Since the region D is a triangular region with vertices (0, 0), (0, 7), and (7, 0), we can express D as D = (x, y) .

The integrand f(x, y) = sin(y). So, the double integral becomes ∫∫Dsin(y)dA.

To evaluate this integral, we need to set up the limits of integration. Since y varies from 0 to 7 - x, and x varies from 0 to 7, we can rewrite the integral as ∫07∫07-xsin(y)dydx.

Integrating with respect to y first, the inner integral becomes -cos(y) evaluated from 0 to 7 - x. This simplifies to (-cos(7 - x)) - (-cos(0)) = -cos(7 - x) + 1.

Now, we integrate with respect to x. The outer integral becomes ∫07(-cos(7 - x) + 1)dx. Integrating, we get -(sin(7 - x)) + x evaluated from 0 to 7. This simplifies to -sin(0) + 7 - (-sin(7)) = 7 - sin(7).

Therefore, the value of the double integral ∬Dsin(y)dA over the region D is 7 - sin(7).