Question

For each of the following iterated integrals, find the region \(R\) and write the integral in the other form (interchanging the order of integration): a) \(\int_{1 / 2}^{1} \int_{0}^{1-x} f(x, y) d y d x\) b) \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y d x\) c) \(\int_{0}^{1} \int_{y-1}^{0} f(x, y) d x d y\) d) \(\int_{0}^{1} \int_{1-x}^{1+x} f(x, y) d y d x\)

Short answer

Answer: The region R can be described as: $$\frac{1}{2} \leq x \leq 1$$ $$0 \leq y \leq 1-x$$ The iterated integral in the other form is: $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) dx dy$$

Step-by-step solution

Identify the limits of integration

For integral a) \(\int_{1 / 2}^{1} \int_{0}^{1-x} f(x, y) d y d x\), we have: - \(x\) ranges from \(1/2\) to \(1\) - \(y\) ranges from \(0\) to \(1-x\)

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$\frac{1}{2} \leq x \leq 1$$ $$0 \leq y \leq 1-x$$

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x \geq 1-y$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) d x d y$$ Answer for a): $$\int_{0}^{1/2} \int_{1-y}^{1} f(x, y) d x d y$$ #b) Find the region R and write the iterated integral in the other form#

Identify the limits of integration

For integral b) \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y d x\), we have: - \(x\) ranges from \(0\) to \(1\) - \(y\) ranges from \(0\) to \(\sqrt{1-x^2}\)

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq x \leq 1$$ $$0 \leq y \leq \sqrt{1-x^2}$$

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x^2 \leq 1 - y^2$$ $$x \leq \sqrt{1 - y^2}$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x d y$$ Answer for b): $$\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} f(x, y) d x d y$$ #c) Find the region R and write the iterated integral in the other form#

Identify the limits of integration

For integral c) \(\int_{0}^{1} \int_{y-1}^{0} f(x, y) d x d y\), we have: - \(y\) ranges from \(0\) to \(1\) - \(x\) ranges from \(y-1\) to \(0\)

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq y \leq 1$$ $$y-1 \leq x \leq 0$$

Interchange the order of integration

To switch the order of integration, we will first solve for \(y\) in the inequality for \(x\): $$y \geq x+1$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{-1}^{0} \int_{x+1}^{1} f(x, y) d y d x$$ Answer for c): $$\int_{-1}^{0} \int_{x+1}^{1} f(x, y) d y d x$$ #d) Find the region R and write the iterated integral in the other form#

Identify the limits of integration

For integral d) \(\int_{0}^{1} \int_{1-x}^{1+x} f(x, y) d y d x\), we have: - \(x\) ranges from \(0\) to \(1\) - \(y\) ranges from \(1-x\) to \(1+x\)

Determine the region R

To find the region R, we will use the limits of integration. The region can be described as: $$0 \leq x \leq 1$$ $$1-x \leq y \leq 1+x$$

Interchange the order of integration

To switch the order of integration, we will first solve for \(x\) in the inequality for \(y\): $$x \leq y-1$$ $$x \geq -y+1$$ Now we can rewrite the iterated integral with the new order of integration: $$\int_{0}^{1} \int_{-y+1}^{y-1} f(x, y) d x d y$$ Answer for d): $$\int_{0}^{1} \int_{-y+1}^{y-1} f(x, y) d x d y$$

Key concepts

Order of Integration

The order of integration refers to the sequence in which we integrate a function of multiple variables. Consider iterated integrals where we perform one integral at a time, typically starting with the innermost integral. When we change the order of integration, it's crucial to take into account the limits and ensure they correctly describe the intended region of integration. For example, in the exercise a), the original order \( dy dx \) suggests that we integrate with respect to \( y \) first, holding \( x \) constant within its bounds. To interchange this order, we re-configure the limits so that \( x \) can be integrated first. This process must be done carefully to maintain the integrity of the region over which we are integrating. The skill of altering the order of integration efficiently is valuable, particularly when one order presents a simpler integral to solve than the other.

Region of Integration

The region of integration, denoted as \( R \), defines the subset of the plane or space over which the function is being integrated. Specifying the correct region is a fundamental step in setting up iterated integrals. It's visualized by the inequalities that are provided by the limits of integration. When changing the order of integration, as we saw in the solution for exercise a), we must ensure that the new limits still represent the same physical region. Graphing these regions on a coordinate system can help demonstrate the bounds on each variable and can be instrumental in understanding the relationships between them. The region of integration for each part of the given exercise is illustrated by inequalities, and the interchangeable order impacts how we describe this region but not its geometry.

Double Integrals

Double integrals extend the concept of a single integral to two dimensions. They are represented as iterated integrals with two different variables. The double integral of a function \( f(x,y) \) over a region \( R \) sums up the values of \( f \) at all points within that region, effectively calculating the 'volume' under the surface described by \( f(x,y) \) over \( R \). The solution steps illustrate how to compute iterated integrals by integrating one variable at a time and showing that the choice of which variable to integrate first can be swapped. For example, in exercise b), we integrate with respect to \( y \) first. However, if we find it more convenient or viable, we can choose to integrate with respect to \( x \) first, as long as we identify the proper limits that correspond to our region \( R \). This flexibility can simplify calculations, especially in cases where one order of integration might be easier to evaluate than the other.