Evaluate The Double Integral Over The Given Region R

Evaluating double integrals over a region R is a fundamental concept in multivariable calculus. It extends the idea of single integration to two dimensions, allowing us to calculate volumes, areas, and other quantities associated with two-dimensional regions. The process involves setting up the integral correctly based on the region's boundaries and then performing iterated integration.

Understanding Double Integrals

A double integral is an integral of a function of two variables over a two-dimensional region. Notationally, it is represented as:

$\iint_R f(x, y) , dA$

Where:

• $f(x, y)$ is the function to be integrated, often representing a height or density.
• $R$ is the region of integration in the xy-plane.
• $dA$ is an infinitesimal area element, which can be expressed as $dx , dy$ or $dy , dx$, depending on the order of integration.

The result of a double integral can be interpreted in different ways:

• If $f(x, y) \geq 0$, the double integral represents the volume under the surface $z = f(x, y)$ and above the region $R$ in the xy-plane.
• If $f(x, y) = 1$, the double integral represents the area of the region $R$.
• If $f(x, y)$ represents a density function, the double integral represents the mass of the region $R$.

Setting Up Double Integrals: Defining the Region R

The key to evaluating a double integral lies in correctly defining the region of integration, $R$. $R$ can be described in two main ways, leading to different integration orders:

1. Type I Region: A region bounded by two vertical lines $x = a$ and $x = b$, and two continuous functions $y = g_1(x)$ and $y = g_2(x)$ such that $g_1(x) \leq g_2(x)$ for all $x$ in $[a, b]$. In this case, the double integral is set up as:

   $\iint_R f(x, y) , dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) , dy , dx$

   We integrate with respect to y first (inner integral), treating x as a constant, and then integrate with respect to x (outer integral).

2. Type II Region: A region bounded by two horizontal lines $y = c$ and $y = d$, and two continuous functions $x = h_1(y)$ and $x = h_2(y)$ such that $h_1(y) \leq h_2(y)$ for all $y$ in $[c, d]$. In this case, the double integral is set up as:

   $\iint_R f(x, y) , dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y) , dx , dy$

   We integrate with respect to x first (inner integral), treating y as a constant, and then integrate with respect to y (outer integral).

Steps for Defining the Region R and Setting Up the Integral:

1. Sketch the Region: Draw a clear diagram of the region R in the xy-plane. This is crucial for visualizing the boundaries.
2. Choose an Order of Integration: Decide whether to integrate with respect to x first (Type II) or y first (Type I). The choice often depends on the shape of the region and the complexity of the function $f(x, y)$. Sometimes, one order will lead to a simpler integral than the other. Consider which order would make the limits of integration easier to determine.
3. Determine the Limits of Integration:
   • If integrating with respect to y first (Type I):
     • Draw a vertical line through the region R. The bottom of the line touches the curve $y = g_1(x)$, and the top touches the curve $y = g_2(x)$. These are the limits of the inner integral.
     • The outer integral limits, a and b, are the smallest and largest x-values that define the region R.
   • If integrating with respect to x first (Type II):
     • Draw a horizontal line through the region R. The left end of the line touches the curve $x = h_1(y)$, and the right end touches the curve $x = h_2(y)$. These are the limits of the inner integral.
     • The outer integral limits, c and d, are the smallest and largest y-values that define the region R.
4. Write the Double Integral: Once you have determined the limits of integration, write out the complete double integral with the correct order of integration and limits.

Evaluating the Iterated Integral

After setting up the double integral, the next step is to evaluate the iterated integral. This involves performing two single integrations, one after the other.

Steps for Evaluating the Iterated Integral:

1. Evaluate the Inner Integral: Treat the outer variable as a constant and integrate the inner integral with respect to the inner variable. For example, if you have:

   $\int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y) , dy , dx$

   First evaluate:

   $F(x) = \int_{g_1(x)}^{g_2(x)} f(x, y) , dy$

   Remember that the result of this integration, F(x), will be a function of x only.

2. Evaluate the Outer Integral: Integrate the result of the inner integral with respect to the outer variable. In the example above, this would be:

   $\int_a^b F(x) , dx$

   The result of this integration will be a single number, which is the value of the double integral.

Important Considerations:

• Order of Integration: Sometimes, the order of integration significantly affects the difficulty of the integral. If one order is too difficult, try switching the order. This involves redefining the region R and finding new limits of integration.
• Constant Limits: If the limits of integration are constants, the order of integration usually doesn't matter (assuming $f(x,y)$ is continuous).
• Techniques of Integration: You may need to use various techniques of integration, such as u-substitution, integration by parts, trigonometric substitution, or partial fractions, to evaluate the inner and outer integrals.

Examples of Evaluating Double Integrals

Let's illustrate the process with some examples:

Example 1: Evaluating a double integral over a rectangular region.

Evaluate the double integral:

$\iint_R (x + y) , dA$

where R is the rectangle defined by $0 \leq x \leq 2$ and $1 \leq y \leq 3$.

Solution:

1. Region: The region R is a rectangle with vertices (0,1), (2,1), (2,3), and (0,3).

2. Order of Integration: Since the limits are constants, we can choose either order. Let's choose to integrate with respect to y first.

3. Limits of Integration:

   • The inner integral will have limits $y = 1$ to $y = 3$.
   • The outer integral will have limits $x = 0$ to $x = 2$.

4. Double Integral:

   $\iint_R (x + y) , dA = \int_0^2 \int_1^3 (x + y) , dy , dx$

5. Evaluate the Inner Integral:

   $\int_1^3 (x + y) , dy = \left[ xy + \frac{y^2}{2} \right]_1^3 = (3x + \frac{9}{2}) - (x + \frac{1}{2}) = 2x + 4$

6. Evaluate the Outer Integral:

   $\int_0^2 (2x + 4) , dx = \left[ x^2 + 4x \right]_0^2 = (4 + 8) - (0) = 12$

Therefore, $\iint_R (x + y) , dA = 12$.

Example 2: Evaluating a double integral over a non-rectangular region (Type I).

Evaluate the double integral:

$\iint_R x , dA$

where R is the region bounded by $y = x^2$ and $y = 4$.

Solution:

1. Region: The region R is bounded above by the line $y = 4$ and below by the parabola $y = x^2$. The curves intersect at $x = -2$ and $x = 2$.

2. Order of Integration: It's easier to integrate with respect to y first (Type I).

3. Limits of Integration:

   • The inner integral will have limits $y = x^2$ to $y = 4$.
   • The outer integral will have limits $x = -2$ to $x = 2$.

4. Double Integral:

   $\iint_R x , dA = \int_{-2}^2 \int_{x^2}^4 x , dy , dx$

5. Evaluate the Inner Integral:

   $\int_{x^2}^4 x , dy = \left[ xy \right]_{x^2}^4 = 4x - x^3$

6. Evaluate the Outer Integral:

   $\int_{-2}^2 (4x - x^3) , dx = \left[ 2x^2 - \frac{x^4}{4} \right]_{-2}^2 = (8 - 4) - (8 - 4) = 0$

Therefore, $\iint_R x , dA = 0$. This result makes sense because the function x is odd and the region is symmetric about the y-axis.

Example 3: Evaluating a double integral over a non-rectangular region (Type II).

Evaluate the double integral:

$\iint_R y , dA$

where R is the region bounded by $x = y^2$ and $x = 2 - y^2$.

Solution:

1. Region: The region R is bounded on the left by the parabola $x = y^2$ and on the right by the parabola $x = 2 - y^2$. The curves intersect when $y^2 = 2 - y^2$, which gives $2y^2 = 2$, so $y = -1$ and $y = 1$.

2. Order of Integration: It's easier to integrate with respect to x first (Type II).

3. Limits of Integration:

   • The inner integral will have limits $x = y^2$ to $x = 2 - y^2$.
   • The outer integral will have limits $y = -1$ to $y = 1$.

4. Double Integral:

   $\iint_R y , dA = \int_{-1}^1 \int_{y^2}^{2 - y^2} y , dx , dy$

5. Evaluate the Inner Integral:

   $\int_{y^2}^{2 - y^2} y , dx = \left[ xy \right]_{y^2}^{2 - y^2} = (2 - y^2)y - y^3 = 2y - 2y^3$

6. Evaluate the Outer Integral:

   $\int_{-1}^1 (2y - 2y^3) , dy = \left[ y^2 - \frac{y^4}{2} \right]_{-1}^1 = (1 - \frac{1}{2}) - (1 - \frac{1}{2}) = 0$

Therefore, $\iint_R y , dA = 0$. Again, this makes sense because the function y is odd and the region is symmetric about the x-axis.

Example 4: Changing the Order of Integration

Evaluate the double integral:

$\int_0^1 \int_x^1 e^{y^2} , dy , dx$

Solution:

1. The Problem: The integral $\int e^{y^2} , dy$ does not have a simple elementary antiderivative. Therefore, we cannot directly evaluate the inner integral.

2. Change the Order of Integration: We need to reverse the order of integration. First, we need to determine the region R.

3. Describe the Region R: The original limits of integration tell us that R is defined by $x \leq y \leq 1$ and $0 \leq x \leq 1$. This is a region in the first quadrant bounded by the lines $y = x$, $y = 1$, and $x = 0$.

4. Redescribe R for Reversed Order: To integrate with respect to x first, we need to describe R as $0 \leq x \leq y$ and $0 \leq y \leq 1$.

5. Reversed Integral: The double integral with the reversed order is:

   $\int_0^1 \int_0^y e^{y^2} , dx , dy$

6. Evaluate the Inner Integral:

   $\int_0^y e^{y^2} , dx = \left[ x e^{y^2} \right]_0^y = y e^{y^2}$

7. Evaluate the Outer Integral:

   $\int_0^1 y e^{y^2} , dy$

   Use u-substitution: let $u = y^2$, then $du = 2y , dy$. So, $\frac{1}{2} du = y , dy$. The limits of integration change to $u = 0$ and $u = 1$.

   $\int_0^1 y e^{y^2} , dy = \frac{1}{2} \int_0^1 e^u , du = \frac{1}{2} \left[ e^u \right]_0^1 = \frac{1}{2} (e - 1)$

Therefore, $\int_0^1 \int_x^1 e^{y^2} , dy , dx = \frac{1}{2} (e - 1)$.

Applications of Double Integrals

Double integrals have numerous applications in various fields:

• Calculating Area: As mentioned earlier, if $f(x, y) = 1$, the double integral gives the area of the region R.
• Calculating Volume: If $f(x, y)$ represents the height of a surface, the double integral gives the volume under the surface and above the region R.
• Finding Mass and Center of Mass: If $f(x, y)$ represents the density of a lamina (a thin plate), the double integral gives the mass of the lamina. The center of mass can also be calculated using double integrals.
• Calculating Moments of Inertia: Double integrals are used to calculate moments of inertia, which are important in physics for describing the resistance of an object to rotational motion.
• Probability and Statistics: Double integrals are used in probability to calculate probabilities associated with continuous random variables.
• Engineering: Double integrals are used in structural analysis, fluid dynamics, and heat transfer.

Common Mistakes to Avoid

• Incorrect Limits of Integration: This is the most common mistake. Always sketch the region R and carefully determine the limits based on the chosen order of integration. Make sure the inner limits are functions of the outer variable.
• Reversing the Order of Integration Incorrectly: When changing the order of integration, make sure to correctly redefine the region R and find the new limits.
• Forgetting the Jacobian: When transforming to a different coordinate system (e.g., polar coordinates), remember to include the Jacobian in the integral. (The Jacobian is a scaling factor that accounts for the distortion of area during the transformation). We haven't discussed this in detail, but it's essential when working with coordinate transformations.
• Arithmetic Errors: Be careful with arithmetic and algebraic manipulations, especially when dealing with complex functions and limits.
• Not Checking for Symmetry: If the function and region are symmetric, the integral might simplify to zero or be easier to calculate. Look for symmetry before diving into the integration.
• Ignoring Discontinuities: Ensure the function f(x, y) is continuous over the region R. If there are discontinuities, the integral may need to be split into multiple integrals.

Conclusion

Evaluating double integrals requires a solid understanding of the region of integration, the ability to set up the integral correctly, and proficiency in single-variable integration techniques. By carefully sketching the region, choosing the appropriate order of integration, and paying attention to detail, you can successfully evaluate double integrals and apply them to solve a wide range of problems in mathematics, physics, engineering, and other fields. Remember to practice regularly and work through various examples to build your skills and intuition. Don't be afraid to change the order of integration if one order leads to a more manageable integral. The ability to correctly set up and evaluate double integrals is a powerful tool in the arsenal of any STEM professional.