Evaluate the double integral (2x – y^2) dA over the triangular region R enclosed between the lines y = -x + 1, y = x + 1, and y = 3. Solve this problem step by step so that I can understand.
The Correct Answer and Explanation is:
Step-by-step Solution:
Step 1: Understand the Region of Integration
The region RRR is bounded by the lines:
- y=−x+1y = -x + 1y=−x+1
- y=x+1y = x + 1y=x+1
- y=3y = 3y=3
Find the points of intersection of the lines:
- y=−x+1y = -x + 1y=−x+1 and y=x+1y = x + 1y=x+1:
Set the equations equal:
−x+1=x+1⇒−x−x=0⇒x=0-x + 1 = x + 1 \Rightarrow -x – x = 0 \Rightarrow x = 0−x+1=x+1⇒−x−x=0⇒x=0
Then y=1y = 1y=1
Point: (0,1)(0, 1)(0,1) - y=−x+1y = -x + 1y=−x+1 and y=3y = 3y=3:
Set:
−x+1=3⇒x=−2-x + 1 = 3 \Rightarrow x = -2−x+1=3⇒x=−2
Point: (−2,3)(-2, 3)(−2,3) - y=x+1y = x + 1y=x+1 and y=3y = 3y=3:
Set:
x+1=3⇒x=2x + 1 = 3 \Rightarrow x = 2x+1=3⇒x=2
Point: (2,3)(2, 3)(2,3)
The triangle has vertices at (−2,3)(-2, 3)(−2,3), (2,3)(2, 3)(2,3), and (0,1)(0, 1)(0,1).
Step 2: Set Up the Integral
The region is best described using vertical slices from y=1y = 1y=1 to y=3y = 3y=3.
At each fixed yyy, xxx goes from the left curve x=−y+1x = -y + 1x=−y+1 to the right curve x=y−1x = y – 1x=y−1.
The double integral becomes:∫y=13∫x=−y+1y−1(2x−y2) dx dy\int_{y=1}^{3} \int_{x=-y+1}^{y-1} (2x – y^2)\, dx\, dy∫y=13∫x=−y+1y−1(2x−y2)dxdy
Step 3: Integrate with Respect to xxx∫13[∫−y+1y−1(2x−y2) dx]dy\int_{1}^{3} \left[ \int_{-y+1}^{y-1} (2x – y^2)\, dx \right] dy∫13[∫−y+1y−1(2x−y2)dx]dy
First, compute the inner integral:∫−y+1y−1(2x−y2) dx=[x2−y2x]x=−y+1x=y−1\int_{-y+1}^{y-1} (2x – y^2)\, dx = \left[ x^2 – y^2 x \right]_{x = -y+1}^{x = y-1}∫−y+1y−1(2x−y2)dx=[x2−y2x]x=−y+1x=y−1
Evaluate at bounds:
- At x=y−1x = y – 1x=y−1:
(y−1)2−y2(y−1)(y – 1)^2 – y^2(y – 1)(y−1)2−y2(y−1) - At x=−y+1x = -y + 1x=−y+1:
(−y+1)2−y2(−y+1)(-y + 1)^2 – y^2(-y + 1)(−y+1)2−y2(−y+1)
Compute:(y−1)2=y2−2y+1,y2(y−1)=y3−y2(y – 1)^2 = y^2 – 2y + 1,\quad y^2(y – 1) = y^3 – y^2(y−1)2=y2−2y+1,y2(y−1)=y3−y2
So the upper limit gives:y2−2y+1−(y3−y2)=−y3+2y2−2y+1y^2 – 2y + 1 – (y^3 – y^2) = -y^3 + 2y^2 – 2y + 1y2−2y+1−(y3−y2)=−y3+2y2−2y+1(−y+1)2=y2−2y+1,y2(−y+1)=−y3+y2(-y + 1)^2 = y^2 – 2y + 1,\quad y^2(-y + 1) = -y^3 + y^2(−y+1)2=y2−2y+1,y2(−y+1)=−y3+y2
So the lower limit gives:y2−2y+1−(−y3+y2)=y3−2y+1y^2 – 2y + 1 – (-y^3 + y^2) = y^3 – 2y + 1y2−2y+1−(−y3+y2)=y3−2y+1
Now subtract:Upper – Lower=(−y3+2y2−2y+1)−(y3−2y+1)=−2y3+2y2\text{Upper – Lower} = (-y^3 + 2y^2 – 2y + 1) – (y^3 – 2y + 1) = -2y^3 + 2y^2Upper – Lower=(−y3+2y2−2y+1)−(y3−2y+1)=−2y3+2y2
Step 4: Integrate with Respect to yyy∫13(−2y3+2y2)dy=−2∫13y3dy+2∫13y2dy\int_{1}^{3} (-2y^3 + 2y^2) dy = -2 \int_{1}^{3} y^3 dy + 2 \int_{1}^{3} y^2 dy∫13(−2y3+2y2)dy=−2∫13y3dy+2∫13y2dy=−2[y44]13+2[y33]13= -2 \left[ \frac{y^4}{4} \right]_{1}^{3} + 2 \left[ \frac{y^3}{3} \right]_{1}^{3}=−2[4y4]13+2[3y3]13
Compute:
- y44∣13=814−14=804=20\frac{y^4}{4} \Big|_1^3 = \frac{81}{4} – \frac{1}{4} = \frac{80}{4} = 204y413=481−41=480=20
- y33∣13=273−13=263\frac{y^3}{3} \Big|_1^3 = \frac{27}{3} – \frac{1}{3} = \frac{26}{3}3y313=327−31=326
Now multiply:−2⋅20+2⋅263=−40+523=−120+523=−683-2 \cdot 20 + 2 \cdot \frac{26}{3} = -40 + \frac{52}{3} = \frac{-120 + 52}{3} = \frac{-68}{3}−2⋅20+2⋅326=−40+352=3−120+52=3−68
Final Answer:−683\boxed{-\frac{68}{3}}−368
Explanation
This double integral problem focuses on evaluating a function over a triangular region defined by intersecting lines. The region of integration is enclosed by three straight lines: y=−x+1y = -x + 1y=−x+1, y=x+1y = x + 1y=x+1, and the horizontal line y=3y = 3y=3. These lines intersect to form a triangle with vertices at (−2,3)(-2, 3)(−2,3), (2,3)(2, 3)(2,3), and (0,1)(0, 1)(0,1).
To simplify the integration, the region is described using vertical slices. For each value of yyy from 1 to 3, the variable xxx ranges from the left boundary x=−y+1x = -y + 1x=−y+1 to the right boundary x=y−1x = y – 1x=y−1. This description allows setting up the limits of the double integral properly.
The inner integral with respect to xxx handles the expression 2x−y22x – y^22x−y2. This expression is integrated with respect to xxx, treating yyy as constant, yielding a result that still includes yyy. Substituting the limits of integration for xxx produces a more complex expression that must be simplified before integrating with respect to yyy.
Once simplified, the resulting single-variable integral in terms of yyy is evaluated using basic power rule techniques. Each term is integrated separately, and definite integrals are calculated using standard limits. Final arithmetic operations combine these results into a single fraction.
The solution reveals the total value of the double integral over the triangular region as −683-\frac{68}{3}−368, showing how area, limits, and algebraic simplification combine in multivariable calculus.
