Math Problem Statement

Evaluate the double integral.

2y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1)

D

Solution

To evaluate the double integral of 2y22y^2 over the triangular region DD with vertices at (0,1)(0, 1), (1,2)(1, 2), and (4,1)(4, 1), we need to determine the limits of integration and set up the integral.

Step 1: Determine the equation of the lines bounding the region

The triangular region has three edges, which can be defined by the equations of the lines passing through each pair of vertices:

  1. Line between (0,1)(0, 1) and (4,1)(4, 1):

    • This is a horizontal line with y=1y = 1.
  2. Line between (1,2)(1, 2) and (4,1)(4, 1):

    • The slope of the line is slope=1241=13.\text{slope} = \frac{1 - 2}{4 - 1} = -\frac{1}{3}.
    • The equation of the line is: y2=13(x1)y=13x+53.y - 2 = -\frac{1}{3}(x - 1) \quad \Rightarrow \quad y = -\frac{1}{3}x + \frac{5}{3}.
  3. Line between (0,1)(0, 1) and (1,2)(1, 2):

    • The slope is slope=2110=1.\text{slope} = \frac{2 - 1}{1 - 0} = 1.
    • The equation of the line is: y1=1(x0)y=x+1.y - 1 = 1(x - 0) \quad \Rightarrow \quad y = x + 1.

Step 2: Set up the double integral

We will integrate with respect to yy first (vertically), and the limits for yy will depend on xx. Specifically:

  • For xx between 0 and 1, yy ranges from y=x+1y = x + 1 to y=2y = 2.
  • For xx between 1 and 4, yy ranges from y=1y = 1 to y=13x+53y = -\frac{1}{3}x + \frac{5}{3}.

The double integral can thus be written as: 01x+122y2dydx+14113x+532y2dydx.\int_0^1 \int_{x+1}^2 2y^2 \, dy \, dx + \int_1^4 \int_1^{-\frac{1}{3}x + \frac{5}{3}} 2y^2 \, dy \, dx.

Step 3: Evaluate the inner integrals

First integral:

01x+122y2dydx.\int_0^1 \int_{x+1}^2 2y^2 \, dy \, dx. The inner integral is: x+122y2dy=2y33x+12=2(23)2((x+1)3)3=162(x+1)33.\int_{x+1}^2 2y^2 \, dy = \frac{2y^3}{3} \Big|_{x+1}^2 = \frac{2(2^3) - 2((x+1)^3)}{3} = \frac{16 - 2(x+1)^3}{3}. Now integrate with respect to xx: 01162(x+1)33dx.\int_0^1 \frac{16 - 2(x+1)^3}{3} \, dx.

Second integral:

14113x+532y2dydx.\int_1^4 \int_1^{-\frac{1}{3}x + \frac{5}{3}} 2y^2 \, dy \, dx. The inner integral is: 113x+532y2dy=2y33113x+53.\int_1^{-\frac{1}{3}x + \frac{5}{3}} 2y^2 \, dy = \frac{2y^3}{3} \Big|_1^{-\frac{1}{3}x + \frac{5}{3}}. Now integrate this with respect to xx.

Step 4: Combine the results

Add the results of both integrals to get the final answer.

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Math Problem Analysis

Mathematical Concepts

Double Integrals
Coordinate Geometry
Integration over Regions

Formulas

Equation of a line: y - y1 = m(x - x1)
Double integral: ∫∫_D f(x, y) dA
Integral of polynomial: ∫ y^n dy = y^(n+1)/(n+1)

Theorems

Fundamental Theorem of Calculus
Area of Triangular Regions

Suitable Grade Level

College Level - Calculus II