Math Problem Statement

Calculate the double integral \( \iint_D f \, dA \) for the given regions \( D \subset \mathbb{R}^2 \) and functions \( f: D \rightarrow \mathbb{R} \).

Solution

The problem you uploaded involves calculating integrals over a given region DD in R2\mathbb{R}^2 and evaluating the integral of a function f(x,y)f(x, y). Here's the problem breakdown:

Problem Statement:

You are asked to calculate the integral DfdA\iint_D f \, dA for the given region DR2D \subset \mathbb{R}^2 and function f:DRf: D \rightarrow \mathbb{R}.

Part (a):

  • Region DD: D={(x,y)R2:x2+y232}D = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 3^2\}, which represents a disk centered at the origin with radius 3.
  • Function f(x,y)f(x, y): f(x,y)=ln(1+x2+y2)f(x, y) = \ln(1 + x^2 + y^2).

Part (b):

  • Region DD: DD is the triangular region with vertices at (0,0)(0,0), (2,0)(2,0), and (0,1)(0,1).
  • Function f(x,y)f(x, y): f(x,y)=exyy2f(x, y) = e^{xy - y^2}.

Approach to Solving the Problem:

  1. Part (a): Since the region is a disk, it is convenient to convert the double integral to polar coordinates. x=rcosθ,y=rsinθx = r\cos\theta, \quad y = r\sin\theta The integral becomes: Dln(1+r2)rdrdθ\iint_D \ln(1 + r^2) \, r \, dr \, d\theta with rr ranging from 0 to 3, and θ\theta ranging from 0 to 2π2\pi.

  2. Part (b): For the triangular region, the limits of integration depend on the vertices of the triangle. The integral can be set up using either dxdydx \, dy or dydxdy \, dx with the appropriate limits determined by the line equations of the triangle.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Double Integrals
Polar Coordinates
Region of Integration

Formulas

Double Integral \( \iint_D f \, dA \)
Transformation to Polar Coordinates \( x = r\cos\theta, \ y = r\sin\theta \)

Theorems

Change of Variables Theorem

Suitable Grade Level

Undergraduate (Calculus II/III)