Evaluate the double integral D is the triangular region with vertices (0,1), (1,2), (4,1).
The Correct Answer and Explanation is:
To evaluate the double integral over the triangular region D with vertices at (0,1),(1,2),(4,1)(0,1), (1,2), (4,1)(0,1),(1,2),(4,1), the region must first be understood and described accurately in terms of boundaries.
Step 1: Find equations of the triangle’s sides
- From (0,1) to (1,2):
This segment has slope m=2−11−0=1m = \frac{2 – 1}{1 – 0} = 1m=1−02−1=1.
Using point-slope form:
y−1=1(x−0)⇒y=x+1y – 1 = 1(x – 0) \Rightarrow y = x + 1y−1=1(x−0)⇒y=x+1 - From (1,2) to (4,1):
Slope is m=1−24−1=−13m = \frac{1 – 2}{4 – 1} = -\frac{1}{3}m=4−11−2=−31
Using point-slope form:
y−2=−13(x−1)⇒y=−13x+73y – 2 = -\frac{1}{3}(x – 1) \Rightarrow y = -\frac{1}{3}x + \frac{7}{3}y−2=−31(x−1)⇒y=−31x+37 - From (0,1) to (4,1):
This is the horizontal line y=1y = 1y=1
Step 2: Set up the integral
Use vertical slices with yyy ranging from 1 to 2. For a fixed yyy, the left and right bounds in xxx must be expressed in terms of yyy:
- Solve y=x+1⇒x=y−1y = x + 1 \Rightarrow x = y – 1y=x+1⇒x=y−1 (left boundary)
- Solve y=−13x+73⇒x=−3y+7y = -\frac{1}{3}x + \frac{7}{3} \Rightarrow x = -3y + 7y=−31x+37⇒x=−3y+7 (right boundary)
So, the integral becomes:∫y=12∫x=y−1−3y+7f(x,y) dx dy\int_{y=1}^{2} \int_{x=y-1}^{-3y+7} f(x,y)\, dx\, dy∫y=12∫x=y−1−3y+7f(x,y)dxdy
If the function is not specified, assume integration of 1 over the region to find the area of the triangle.∫12∫y−1−3y+71 dx dy\int_{1}^{2} \int_{y-1}^{-3y+7} 1\, dx\, dy∫12∫y−1−3y+71dxdy
Step 3: Evaluate inner integral∫12[−3y+7−(y−1)] dy=∫12[−4y+8] dy\int_{1}^{2} [-3y + 7 – (y – 1)]\, dy = \int_{1}^{2} [-4y + 8]\, dy∫12[−3y+7−(y−1)]dy=∫12[−4y+8]dy
Now integrate:∫12(−4y+8) dy=[−2y2+8y]12\int_{1}^{2} (-4y + 8)\, dy = [-2y^2 + 8y]_{1}^{2}∫12(−4y+8)dy=[−2y2+8y]12=[−2(4)+16]−[−2(1)+8]=[−8+16]−[−2+8]=8−6=2= [-2(4) + 16] – [-2(1) + 8] = [-8 + 16] – [-2 + 8] = 8 – 6 = 2=[−2(4)+16]−[−2(1)+8]=[−8+16]−[−2+8]=8−6=2
Final Answer:2\boxed{2}2
Explanation (Approx. 300 words):
The problem involves computing a double integral over a triangular region defined by three points. To evaluate it properly, one must first identify the region’s shape and the lines forming its sides. Connecting the points (0,1)(0,1)(0,1), (1,2)(1,2)(1,2), and (4,1)(4,1)(4,1), the region forms a triangle. Each line segment corresponds to a linear equation derived from the slope and one of the points.
Once the boundaries are determined, the next step involves expressing the region in terms of an integral. Choosing vertical slices allows expressing xxx as a function of yyy, from the left side of the triangle to the right side. The lower limit of xxx is given by the line from (0,1)(0,1)(0,1) to (1,2)(1,2)(1,2), and the upper limit by the line from (1,2)(1,2)(1,2) to (4,1)(4,1)(4,1).
Because no specific function is given, the integral evaluates the area of the triangular region. The inner integral calculates the horizontal distance across the triangle at a fixed height yyy. The result of this integration yields a function of yyy, which is then integrated vertically from the bottom to the top of the triangle.
After evaluating both integrals, the result shows the area enclosed by the triangle equals 2 square units.
