Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Write but do not compute the iterated form of the integral \(\int_{R} \int F(P) d A\) for the functions \(F\) and domains indicated. In i. and j. write the integral as the sum of two iterated integrals. a. \(\quad F(x, y)=x \times y \quad 0\) \(\leq x \quad \leq 3 \quad 0 \leq y \leq 2\) b. \(\quad F(x, y)=x+y \quad 1 \quad \leq x \quad \leq 3 \quad 2 \leq y \leq 5\) c. \(\quad F(x, y)=x \times \ln y \quad 0 \quad \leq x \quad \leq 3 \quad 1 \leq y \leq 3\) d. \(\quad F(x, y)=x^{2} y\) 0\(\leq x \quad \leq \pi \quad 0 \leq y \leq \sin x\) e. \(\quad F(x, y)=x+y \quad 1 \quad \leq x\) \(x \leq y \leq x^{2}\) f. \(\quad F(x, y)=x \times y^{2} \quad 1 \quad \leq y \quad \leq 2 \quad y \leq x \leq y^{2}\) g. \(\quad F(x, y)=x \times y \quad 0 \quad \leq x+y \quad \leq 2 \quad 0 \leq x, \quad 0 \leq y\) h. \(\quad F(x, y)=x+y \quad 0 \leq x^{2}+y^{2} \leq 1\) i. \(\quad F(x, y)=x \times \ln y \quad 1 \quad \leq x+y \quad \leq 3 \quad 0 \leq x, \quad 0 \leq y\) j. \(F(x, y)=x \times \ln y \quad 1 \leq x^{2}+y^{2} \leq 4 \quad 0 \leq x, \quad 0 \leq y\)

Step-by-Step Solution

Answer

Set each integral as an iterated integral based on boundary conditions for each case.

Step 1: Understanding Iterated Integrals

For iterated integrals, we need to express a double integral \[ \int_{R} \int F(x, y) \, dx \, dy \] in terms of two single integrals. We choose the order of integration based on the boundaries.

Step 2: Setting Up the Integral for Part a

For the function \( F(x,y) = x \times y \), with bounds \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 2 \), the iterated integral can be written in two ways: \[ \int_{0}^{2} \int_{0}^{3} x \times y \, dx \, dy \] or \[ \int_{0}^{3} \int_{0}^{2} x \times y \, dy \, dx. \]

Step 3: Setting Up the Integral for Part b

For the function \( F(x,y) = x + y \) with bounds \( 1 \leq x \leq 3 \) and \( 2 \leq y \leq 5 \), the integral can be set up as: \[ \int_{2}^{5} \int_{1}^{3} (x + y) \, dx \, dy \] or \[ \int_{1}^{3} \int_{2}^{5} (x + y) \, dy \, dx. \]

Step 4: Setting Up the Integral for Part c

For the function \( F(x,y) = x \ln y \) with bounds \( 0 \leq x \leq 3 \) and \( 1 \leq y \leq 3 \), the iterated integral is: \[ \int_{1}^{3} \int_{0}^{3} x \ln y \, dx \, dy \] or \[ \int_{0}^{3} \int_{1}^{3} x \ln y \, dy \, dx. \]

Step 5: Setting Up the Integral for Part d

For the function \( F(x,y) = x^2 y \) with bounds \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \sin x \), the iterated integral is: \[ \int_{0}^{\pi} \int_{0}^{\sin x} x^2 y \, dy \, dx. \] We can only use this order because the upper limit for \( y \) is a function of \( x \).

Step 6: Setting Up the Integral for Part e

For the function \( F(x,y) = x + y \) with bounds \( 1 \leq x \) and \( x \leq y \leq x^2 \), the iterated integral can be set as: \[ \int_{1}^{\infty} \int_{x}^{x^2} (x + y) \, dy \, dx. \] Again, this order must be used because the \( y \) bounds depend on \( x \).

Step 7: Setting Up the Integral for Part f

For the function \( F(x,y) = x \times y^2 \) with bounds \( 1 \leq y \leq 2 \) and \( y \leq x \leq y^2 \), we can express as: \[ \int_{1}^{2} \int_{y}^{y^2} x \times y^2 \, dx \, dy. \] This order is necessary since \( x \) bounds depend on \( y \).

Step 8: Setting Up the Integral for Part g

For \( F(x,y) = x \times y \) within the region \( 0 \leq x+y \leq 2 \), both \( x, y \geq 0 \), the iterated integral can be split at \( x+y=c \) for values within the triangle. However, integration in conventional form is complex, thus needs more specific segmentation based on limits.

Step 9: Setting Up the Integral for Part h

For \( F(x,y) = x + y \) with the disk \( 0 \leq x^2 + y^2 \leq 1 \), a polar coordinate system is more suitable due to radial symmetry.

Step 10: Setting Up the Integral for Part i

For \( F(x,y) = x \ln y \) with the region \( 1 \leq x+y \leq 3 \), the iterated integral is done separately across the line boundary: \[ \int_{A} + \int_{B}. \] Specific consideration of intersection lines due to changing boundaries.

Step 11: Setting Up the Integral for Part j

For \( F(x,y) = x \ln y \) in the annular region \( 1 \leq x^2 + y^2 \leq 4 \), express the integral as the difference between two circles, achievable by polar coordinates.Each subregion can be iterated separately based on changing boundary dynamics.

Key Concepts

Iterated Integration

Iterated integration involves solving double integrals by reducing them into two successive single integrals. This technique simplifies the computation of multi-variable integrals. Instead of evaluating the integral over a region directly, we integrate one variable at a time.

For instance, consider a double integral of the function \( F(x, y) = x \times y \) over a rectangular region \( R \). The process involves choosing an order—for example, integrating with respect to \( x \) first, then \( y \). This is written as:

• \[ \int_{0}^{2} \left( \int_{0}^{3} x \times y \, dx \right) \, dy \]

Alternatively, you can swap the order of integration. Integrate first with respect to \( y \) and then \( x \):

• \[ \int_{0}^{3} \left( \int_{0}^{2} x \times y \, dy \right) \, dx \]

The beauty of iterated integrals is in their flexibility and utility, allowing for different computational strategies depending on the region and the function involved.

Bounds of Integration

The bounds of integration define the limits within which we evaluate the integral. They can be constants or functions, delineating the region over which we integrate.

For example, for the function \( F(x, y) = x \times y \) over the region defined by \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 2 \), these bounds describe a rectangle in the \( xy \)-plane.

In more complex situations, bounds may depend on another variable. Consider function \( F(x, y) = x^2 y \) with bounds \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \sin x \). Here, \( y \)'s upper limit directly depends on \( x \), indicating a varying boundary. Such dependency dictates that integration must be performed in a specific order, which will satisfy the bounds as variable limits change throughout the region.

Understanding these bounds is crucial since they fundamentally define the area or volume over which the integration occurs.

Function of Two Variables

A function of two variables, \( f(x, y) \), assigns a unique real number to each pair \( (x, y) \) in its domain. These functions create a surface in three-dimensional space, where each function value represents a height above or below the \( xy \)-plane.

Such functions are common in many fields, modeling elevation, temperature, and more. For instance, in the problem where \( F(x, y) = x \times y \), each point \( (x, y) \) generates a height \( x \times y \).

Functions like \( F(x, y) = x \log y \) introduce additional complexity due to logarithmic components. These functions emphasize how multiplying basic functions can enrich the behavior and analysis of surfaces. Recognizing patterns and relations in these functions is key in setting up accurate integrals for computation.

Order of Integration

The order of integration refers to the sequence in which multiple integrations are performed. It's a choice that can simplify or complicate the integral, depending on the boundaries and the function itself.

Let's say we have \( F(x, y) = x \times y^2 \) with bounds \( 1 \leq y \leq 2 \) and \( y \leq x \leq y^2 \). To capture the region accurately, we should integrate first with respect to \( x \), because \( x \)'s bounds are given as functions of \( y \).

If the bounds are simple constants, like \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq 2 \), either order may be used. However, when bounds depend on each other, the appropriate order is crucial:

• First: Integrate with respect to the variable with limits that do not depend on the other.
• Next: Integrate the resultant expression with respect to the remaining variable.

Choosing the correct order of integration efficiently handles complex boundaries, reducing the potential for errors.