Question

For each of the following choice of \(R\), represent \(\iint f(x, y) d A\) over \(R\) as an iterated integral of both forms (4.33) and (4.34): a) \(1 \leq x \leq 2,1-x \leq y \leq 1+x\) b) \(y^{2}+x(x-1) \leq 0\)

Short answer

Question: For each region, set up the double integral of the function f(x, y) in both forms 4.33 and 4.34. Answer: For Region a (1 ≤ x ≤ 2, 1-x ≤ y ≤ 1+x): - Form 4.33: \(\int_{1}^{2}\int_{1-x}^{1+x} f(x, y) dy dx\) - Form 4.34: \(\int_{0}^{1}\int_{y-1}^{-y+1} f(x, y) dx dy\) For Region b (\(y^2 + x(x-1) \leq 0\)): - Form 4.33: \(\int_{0}^{1}\int_{-\sqrt{1/4-(x-\frac{1}{2})^2}}^{\sqrt{1/4-(x-\frac{1}{2})^2}} f(x, y) dy dx\) - Form 4.34: \(\int_{-1/2}^{1/2}\int_{\frac{1}{2}-\sqrt{1/4-y^2}}^{\frac{1}{2}+\sqrt{1/4-y^2}} f(x, y) dx dy\)

Step-by-step solution

Identify the region in the xy-plane

The given region is a trapezoid, bounded by the lines \(x=1\), \(x=2\), \(y=1-x\), and \(y=1+x\). We will be integrating a function \(f(x,y)\) over this region.

Set up the double integral as an iterated integral in the form 4.33

In this form, we first integrate w.r.t. y and then w.r.t. x. The bounds for y are the lines \(y=1-x\) and \(y=1+x\), and the bounds for x are \(x=1\) and \(x=2\). The iterated integral in this form is: \[\int_{1}^{2}\int_{1-x}^{1+x} f(x, y) dy dx\]

Set up the double integral as an iterated integral in the form 4.34

In this form, we first integrate w.r.t. x and then w.r.t. y. If we rearrange the inequalities, we see that bounds for x are the lines \(x=y-1\) and \(x=-y+1\). The upper and lower bounds for y can be found by solving \(1-x = 1+x\), which results in the single value \(y=1\). The iterated integral in this form is: \[\int_{0}^{1}\int_{y-1}^{-y+1} f(x, y) dx dy\] #Region b: y^{2} + x(x-1) \leq 0#

Identify the region in the xy-plane

The given inequality \(y^2 + x(x-1) \leq 0\) can be written as \((x-\frac{1}{2})^2+y^2\geq\frac{1}{4}\). This describes a region outside a circle centered at \((1/2,0)\) with radius 1/2.

Set up the double integral as an iterated integral in the form 4.33

The bounds for y with respect to x in this region are given by \(\pm\sqrt{1/4 -(x-\frac{1}{2})^2}\). The bounds for x can be found from the intersections of circle with x-axis, which are \(x=0\) and \(x=1\). The iterated integral for this form is: \[\int_{0}^{1}\int_{-\sqrt{1/4-(x-\frac{1}{2})^2}}^{\sqrt{1/4-(x-\frac{1}{2})^2}} f(x, y) dy dx\]

Set up the double integral as an iterated integral in the form 4.34

The bounds for x with respect to y in this region are given by the circle equations, or \(x=\frac{1}{2} \pm \sqrt{1/4-y^2}\). The bounds for y are found from the intersections of the circle with y-axis, which are \(y=-1/2\) and \(y=1/2\). The iterated integral for this form is: \[\int_{-1/2}^{1/2}\int_{\frac{1}{2}-\sqrt{1/4-y^2}}^{\frac{1}{2}+\sqrt{1/4-y^2}} f(x, y) dx dy\]

Key concepts

Double Integral

A double integral is a mathematical expression representing the aggregate of a function over a two-dimensional area. It's an extension of the concept of a single integral, which represents the summation of a function across an interval. In the context of advanced calculus, double integrals, denoted as \(\iint f(x, y) dA\), are used to compute the volume under a surface defined by the function \(f(x, y)\) above a region in the xy-plane.

For example, in the provided exercise, the function \(f(x, y)\) is to be integrated over a specific region \(R\). To compute the double integral, we translate this two-dimensional problem into two one-dimensional integrations, carried out one inside the other - hence the term 'iterated integral'.

Furthermore, it is crucial to decide the order of integration, which affects the limits or bounds for each integral. Recognizing the appropriate bounds of integration and evaluating the integral sequentially leads to the desired result: the accumulation of the function's values over the prescribed region.

Integration Bounds

The boundaries within which a function is integrated, known as integration bounds, are essential to solving a double integral. The bounds can be either constants, indicating a rectangular region of integration, or variable expressions, reflecting a more complex region.

For the example at hand, in Form 4.33, we see that the integration bounds for \(y\) are dependent on \(x\), ranging from \(1-x\) to \(1+x\). This implies that for every value of \(x\) in the region between 1 and 2, the value of \(y\) will vary, outlining the trapezoidal area over which \(f(x, y)\) is to be integrated.

In Form 4.34, conversely, the bounds are flipped. Here, the outer integral's variable is \(y\), and the inner integral's bounds are in terms of \(y\), i.e., \(y-1\) to \(1-y\). Understanding how to express these bounds correctly is crucial for the proper evaluation of the double integral.

Region of Integration

Identifying the region of integration is a pivotal step in setting up an iterated integral. The region of integration \(R\) encapsulates the domain over which the function \(f(x, y)\) is integrated. This region must be clearly defined and comprehended to correctly establish the limits for the double integral.

In simpler terms, it is the 'shape' over which we are 'painting' the function values to accumulate the sum. For instance 1, the region is a trapezoid in the xy-plane, confined by the lines \(x=1\), \(x=2\), \(y=1-x\), and \(y=1+x\).

Instance 2 involves a region outside a semi-circle, requiring a different approach to set the bounds for the iterated integral. Henceforth, the ability to sketch or visualize this region impacts the efficiency with which one can perform the integration.

Advanced Calculus

Advanced calculus, also known as multivariable calculus, delves into concepts that extend beyond the single-variable calculus. These include multiple integrals (double, triple, etc.), partial derivatives, and more complex geometrical regions in higher dimensions.

In advanced calculus, mastering the art of integrating over different shapes such as rectangles, circles, and arbitrary curves is a cornerstone skill. Recognizing when to apply techniques such as substitution or polar coordinates is pivotal. When faced with double integrals, one must be nimble in deciphering the given region and adept at choosing the appropriate order of integration to simplify the process.

It is also in this realm where visualizing and sketching the region of integration becomes indispensable. Ultimately, advanced calculus empowers students to solve more complex real-world problems by giving them the tools to calculate over varied terrains, whether in engineering, physics, or economics.