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Chapter 14: Problem 38

Show that if \(f(x, y)\) is constant on the rectangle \(R=[a, b] \times[c, d],\) say \(f(x, y)=k,\) then \(f_{\text {ave }}=k\) over \(R .\)

Short Answer

Expert verified
The average value \(f_{\text{ave}}\) is \(k\).

Step by step solution

01

Understand the Problem

We need to show that if a function \(f(x, y)\) is constant over a rectangle, then the average value of the function over that rectangle is equal to the constant value. Given that \(f(x, y) = k\) for all points in the rectangle \(R = [a, b] \times [c, d]\), our task is to prove that the average value of this function over \(R\) is \(k\).
02

Recall the Formula for Average Value of a Function

The average value \(f_{\text{ave}}\) of a function \(f(x, y)\) over a rectangle \(R = [a, b] \times [c, d]\) is calculated by the formula:\[ f_{\text{ave}} = \frac{1}{(b-a)(d-c)} \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx. \]
03

Substitute the Constant Function into the Integral

Since \(f(x, y) = k\) everywhere on \(R\), we substitute \(k\) for \(f(x, y)\) into the formula for the average value:\[ f_{\text{ave}} = \frac{1}{(b-a)(d-c)} \int_{a}^{b} \int_{c}^{d} k \, dy \, dx. \]
04

Solve the Integral

The function inside the integral is constant, so the integral simplifies:\[ \int_{c}^{d} k \, dy = k(d-c). \]Substituting back into the outer integral:\[ \int_{a}^{b} k(d-c) \, dx = k(d-c)(b-a). \]
05

Calculate the Average Value

Now, substitute back into the average value formula:\[ f_{\text{ave}} = \frac{1}{(b-a)(d-c)} \times k(d-c)(b-a). \]This simplifies to:\[ f_{\text{ave}} = k. \]
06

Final Step: Conclusion

We have shown that the average value \(f_{\text{ave}}\) is indeed the constant \(k\), verifying our original statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Function
A constant function is a special type of function where the output value remains the same across its entire domain. This means that no matter the input values, the function will always return the same result:

\[ f(x, y) = k \]
where \( k \) is a fixed constant.
  • Such functions are simple because they don't change; they are always equal to the constant everywhere.
  • For a two-variable function \( f(x, y) \), if it is constant on a rectangular region \( R \), it implies \( f(x, y) = k \) for every point within that rectangle.
This feature of being unchanging makes dealing with constant functions quite straightforward, especially when calculating the average value over any region.
Double Integral
The double integral is a way to integrate over a two-dimensional area. It is particularly useful when calculating the total accumulation of a quantity, such as area under a surface:

\[ \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx \]
  • The process involves performing two integrations - one with respect to \( y \) and another with respect to \( x \).
  • This helps in finding areas, volumes, and average values involving two-dimensional regions.
When applied to a constant function over a particular region, the integration process becomes markedly simpler.
Simplifying becomes effortless because the integral of a constant is just the product of the constant and the length over which it is integrated.
Rectangular Region
A rectangular region in the context of integrals is a defined two-dimensional area, typically aligned with the coordinate axes. It is described using intervals for each variable:

\[ R = [a, b] \times [c, d] \]
  • Here, \([a, b]\) represents the range of \( x \)-values, and \([c, d]\) represents the range of \( y \)-values.
  • It forms a rectangle on the x-y plane with the corners at \((a, c), (a, d), (b, c), \) and \((b, d)\).
Working with such regions is quite straightforward in calculations and illustrations of functions or integrals.
The regularity, with sides parallel to axes, simplifies many mathematical operations like finding average values or calculating area and volume.

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Most popular questions from this chapter

Use cylindrical coordinates. Find the mass of the solid with density \(\delta(x, y, z)=3-z\) that is bounded by the cone \(z=\sqrt{x^{2}+y^{2}}\) and the plane \(z=3\)

(a) Verify that $$ \left|\begin{array}{ll}{a_{1}} & {b_{1}} \\ {c_{1}} & {d_{1}}\end{array}\right|\left|\begin{array}{ll}{a_{2}} & {b_{2}} \\ {c_{2}} & {d_{2}}\end{array}\right|=\left|\begin{array}{ll}{a_{1} a_{2}+b_{1} c_{2}} & {a_{1} b_{2}+b_{1} d_{2}} \\ {c_{1} a_{2}+d_{1} c_{2}} & {c_{1} b_{2}+d_{1} d_{2}}\end{array}\right| $$ (b) If \(x=x(u, v), y=y(u, v)\) is a one-to-one transformation, then \(u=u(x, y), v=v(x, y) .\) Assuming the necessary differentiability, use the result in part (a) and the chain rule to show that $$ \frac{\partial(x, y)}{\partial(u, v)} \cdot \frac{\partial(u, v)}{\partial(x, y)}=1 $$

Find the mass and center of gravity of the lamina. A lamina with density \(\delta(x, y)=x^{2}+y^{2}\) is bounded by the \(x\) -axis and the upper half of the circle \(x^{2}+y^{2}=1\)

The tendency of a solid to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If the solid occupies a region \(G\) in an \(x y z\) -coordinate system, and if its density function \(\delta(x, y, z)\) is continuous on \(G,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iiint_{G}\left(y^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{y} &=\iiint_{G}\left(x^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{z} &=\iiint_{G}\left(x^{2}+y^{2}\right) \delta(x, y, z) d V \end{aligned}$$ In these exercises, find the indicated moments of inertia of the solid, assuming that it has constant density \(\delta .\) \(I_{z}\) for the solid cylinder \(x^{2}+y^{2} \leq a^{2}, 0 \leq z \leq h\)

The transformation \(x=a u, y=b v(a>0, b>0)\) can be rewritten as \(x / a=u, y / b=v,\) and hence it maps the circular region $$ u^{2}+v^{2} \leq 1 $$ into the elliptical region $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 $$ In these exercises, perform the integration by transforming the elliptical region of integration into a circular region of integration and then evaluating the transformed integral in polar coordinates. \(\iint_{R} e^{-\left(x^{2}+4 y^{2}\right)} d A,\) where \(R\) is the region enclosed by the ellipse \(\left(x^{2} / 4\right)+y^{2}=1\)
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