Question

Sketch the regions of integration and evaluate the following integrals. \(\iint_{R}(x+y) d A ; R\) is bounded by \(y=|x|\) and \(y=4\)

Short answer

) - (-x^2/2 + x^2)] = [x^2 + x^2] = 2x^2$$ #tag_title# Step 4: Evaluate the outer integral with respect to y #tag_content# Integrate the expression 2x^2 with respect to y: $$\int_{-4}^{4} 2x^2 dy = 2x^2y\Big|_{-4}^{4}$$ Now, evaluate at the bounds: $$2x^2(4) - 2x^2(-4) = 16x^2 + 16x^2 = 32x^2$$ Since x = ±4, the answer remains constant for both values of x. Therefore, the double integral evaluates to: $$32x^2 = 32(4)^2 = 32(16) = 512$$ So, the final answer is: $$\int\int_R (x + y) dA = 512$$

Step-by-step solution

Sketch the region of integration R

Begin by plotting the given functions y = |x| and y = 4 on a coordinate plane. The function y = |x| is comprised of two lines y = x and y = -x, each of which form a 45-degree angle with the x-axis. To find the points of intersection of y = |x| and y = 4, solve for x when y = 4: $$x = \pm 4$$ The region of integration R is the area enclosed between the lines y = x, y = -x, and y = 4.

Convert the double integral to a pair of single integrals

Express the given double integral as an iterated integral, integrating first with respect to x and then with respect to y: $$\int_{-4}^{4} \int_{-x}^{x} (x + y) dxdy$$

Evaluate the inner integral with respect to x

Integrate the expression (x + y) with respect to x: $$\int_{-x}^{x} (x + y) dx = [x^2/2 + x y]_{-x}^{x}$$ Now, evaluate at the bounds: $$[(x^2/2 + x^2

Key concepts

Regions of Integration

In the context of double integrals, the concept of "regions of integration" refers to the area over which we will integrate in the xy-plane. It is essential to accurately define this region to ensure that we are integrating over the correct space.
To determine the regions of integration, we need to analyze the boundaries given by the functions involved. For the given exercise, the region of integration is defined by the functions \(y = |x|\) and \(y = 4\).

Here's how you approach identifying these regions:

• The function \(y = |x|\) consists of two lines: \(y = x\) and \(y = -x\). These lines intersect the horizontal line \(y = 4\) at points \((4, 4)\) and \((-4, 4)\), respectively. So the boundary in the y-direction is from 0 to 4.
• For any given \(y\) between 0 and 4, \(x\) varies between \(-y\) and \(+y\), capturing the region between the lines until it meets \(y = 4\).

The region of integration (R) is the triangular area enclosed by these boundaries—where the lines \(y = x\), \(y = -x\), and \(y = 4\) meet.

Iterated Integrals

Iterated integrals are a method used to calculate double integrals by breaking them down into a sequence of single integrals. This technique simplifies the process of integration over complex regions by iterating through one variable at a time.
For the integral \(\iint_R (x+y) \, dA\), tackling this in an iterated format involves setting up the integration over two variables sequentially:

Steps to form iterated integrals:

• First, decide the order of integration. In our case, we integrate first with respect to \(x\) and then \(y\).
• The integral in our exercise, expressed as iterated integrals, becomes \[ \int_{0}^{4} \int_{-y}^{y} (x + y) \, dx \, dy \].

The order of integration can affect the difficulty of evaluation and is dictated by the region's geometry. Understanding the limits for each variable is crucial to correctly setting up the bounds for integration. Once determined, evaluate the inner integral first, followed by the outer integral to find the final value.

Coordinate Plane Sketching

Sketching on the coordinate plane helps visualize areas, functions, and intersections critical for solving integrals. For integrals involving regions like in our exercise, sketching is a starting point that leads to better understanding and accurate setup of the integral's bounds.
Here's how to effectively sketch for our specific problem:

• Plot the line \(y = |x|\). This involves drawing \(y = x\) and \(y = -x\) emanating from the origin into the first and second quadrants respectively.
• Draw the horizontal line \(y = 4\), which intersects \(y = x\) and \(y = -x\) at \(x = 4\) and \(x = -4\).
• The region of integration \(R\) is the triangle formed by these lines. It's enclosed by the portions of \(y = x\), \(y = -x\), extending vertically at \(y = 4\).

Proper sketches help in estimating bounds quickly and prevent mistakes in setting up the integrals. They give us spatial awareness that equations alone may not readily provide. Thus, sketching is an indispensable tool in solving double integrals.