Calculate double integral over the triangular region D, with vertices (0, 0), (1, 2) and (0,3).
The Correct Answer and Explanation is:
To evaluate the double integral ∬_D f(x, y) dA over the triangular region with vertices (0, 0), (1, 2), and (0, 3), first identify the boundaries.
The triangle lies between:
- the vertical line x = 0 (since two vertices lie on it),
- the line joining (0, 0) to (1, 2), which has slope 2 ⇒ y = 2x,
- the line joining (0, 3) to (1, 2), which has slope –1 ⇒ y = –x + 3.
This triangle is bounded between y = 2x and y = –x + 3, for x in the interval [0, 1].
Assume the function being integrated is f(x, y) = 1 (to calculate the area of the region, a common choice when no function is specified). The double integral becomes:∫x=01∫y=2x−x+31 dy dx\int_{x=0}^{1} \int_{y=2x}^{-x+3} 1\,dy\,dx∫x=01∫y=2x−x+31dydx
Step 1: Integrate with respect to y:∫y=2x−x+31 dy=[−x+3−2x]=−3x+3\int_{y=2x}^{-x+3} 1\,dy = [-x+3 – 2x] = -3x + 3∫y=2x−x+31dy=[−x+3−2x]=−3x+3
Step 2: Integrate with respect to x:∫x=01(−3x+3) dx=∫01−3x dx+∫013 dx=[−3×22]01+[3x]01=(−32)+3=32\int_{x=0}^{1} (-3x + 3)\,dx = \int_{0}^{1} -3x\,dx + \int_{0}^{1} 3\,dx = \left[-\frac{3x^2}{2}\right]_0^1 + [3x]_0^1 = \left(-\frac{3}{2}\right) + 3 = \frac{3}{2}∫x=01(−3x+3)dx=∫01−3xdx+∫013dx=[−23×2]01+[3x]01=(−23)+3=23
Final Answer:
32\boxed{\frac{3}{2}}23
Explanation (300 words):
To determine the value of the double integral over a triangular region, analyze the shape defined by its vertices. The triangle with points (0, 0), (1, 2), and (0, 3) lies within the first quadrant. Two of the points share the same x-coordinate (x = 0), indicating a vertical side. The third point, (1, 2), forms slanted sides when connected to the other two.
The line from (0, 0) to (1, 2) rises steeply, producing the equation y = 2x. The line from (0, 3) down to (1, 2) slopes downward, yielding y = –x + 3. These two lines intersect at x = 1. Hence, the region bounded between these curves from x = 0 to x = 1 forms the triangle.
Setting up a double integral to find the area of this triangle involves integrating the function f(x, y) = 1 over the domain D. This reflects the concept that the double integral of 1 over a region equals its area. The inner integral represents the change in y from the lower curve (y = 2x) to the upper curve (y = –x + 3), and the outer integral covers x from 0 to 1.
Evaluating the inner integral yields a linear function in terms of x. The outer integral then involves a straightforward polynomial integration. Simplifying the result gives a final area of 3/2 square units. This confirms the area of the triangular region bounded by the given lines and vertices.
