Problem 28
Question
If \(y = f ( x )\) is a nonnegative continuous function over the closed interval \(a \leq x \leq b ,\) show that the double integral definition of area for the closed plane region bounded by the graph of \(f ,\) the vertical lines \(x = a\) and \(x = b ,\) and the \(x\) -axis agrees with the definition for area beneath the curve in Section \(5.3 .\)
Step-by-Step Solution
Verified Answer
The area found using the double integral \( \int_{a}^{b} f(x) \ dx \) matches the definition of area under the curve from Section 5.3.
1Step 1: Understanding the Double Integral Definition of Area
The double integral for the area under a curve over a region is given by \( \iint_{R} 1 \ dA \), where \( R \) is the region beneath the curve of \( y = f(x) \), between \(x = a\) and \(x = b\). Thus, this integral equals \( \int_{a}^{b} \int_{0}^{f(x)} 1 \ dy \ dx \). This involves integrating with respect to \( y \), then \( x \).
2Step 2: Evaluate the Inner Integral
The inner integral \( \int_{0}^{f(x)} 1 \ dy \) results in \( f(x) \), because the integral of \(1\) with respect to \(y\) from \(0\) to \(f(x)\) is the length of the interval \([0, f(x)]\).
3Step 3: Evaluate the Outer Integral
Now, the outer integral becomes \( \int_{a}^{b} f(x) \ dx \). Integrating \(f(x)\) with respect to \(x\) over \([a, b]\) gives the area under the curve as defined in Section 5.3.
4Step 4: Conclusion
Both the double integral and the single variable integral definitions yield the same area \( \int_{a}^{b} f(x) \ dx \), confirming the equivalence of the two definitions for the area under a curve.
Key Concepts
Double IntegralArea Under a CurveContinuous FunctionIntegralsThomas Calculus
Double Integral
A double integral is a fundamental concept in calculus, especially when dealing with functions of two or more variables. It extends the idea of an integral to higher dimensions. In the original exercise, we use a double integral to find the area beneath a function. This involves two integration processes.
- First, it's performed over the variable that corresponds to the height of the area (usually the y-axis).
- Then, it's done over the base of that area, typically the x-axis, to cover the region fully.
- areas of complex surfaces,
- volumes under surfaces, and
- averages over regions.
Area Under a Curve
Finding the area under a curve is a classical problem in calculus. It helps in understanding quantities like total distance or total accumulated change. The area under the curve represents the integral of a function over a specified interval.
- In a single integral, it calculates the area beneath the function graph directly, as seen in the formula \( \int_{a}^{b} f(x) \ dx \).
- In the context of the double integral from our problem, it first calculates the area vertically (inner integral along y) and then horizontally (outer integral along x).
Continuous Function
A continuous function is fundamental in calculus because it guarantees the function behaves predictably over an interval. Continuity in mathematical terms means
- there are no breaks, jumps, or holes in the graph of the function,
- and the function can be drawn without lifting the pen off the paper.
- the integral will have a well-defined value over the interval \([a, b]\),
- and that definitions using single or double integrals remain consistent.
Integrals
Integrals are a core part of calculus, representing the accumulation of quantities and the notion of adding up infinitesimally small pieces. In this context, integrating a function involves:
- breaking down the problem into small parts (infinitesimals), and
- adding them together to understand the whole.
- solve problems involving rates of change by finding total accumulated values,
- are used in physics for concepts like work and energy, and
- provide the necessary tools for analyzing curves and surfaces in more complex, multi-dimensional spaces.
Thomas Calculus
Thomas Calculus is a popular textbook among students learning calculus. It provides a thorough introduction to the foundational concepts, including integrals and their applications.
In this context:
- Thomas Calculus helps explain the equivalence of different definitions for the area under a curve, as shown in sections like 5.3.
- It emphasizes visual understanding through graphs and intuitive explanations.
- By using established examples and exercises, students can connect the theoretical with the practical, seeing how integrals operate in real-world problems.
Other exercises in this chapter
Problem 28
Cardioid overlapping a circle Find the area of the region that lies inside the cardioid \(r=1+\cos \theta\) and outside the circle \(r=1\)
View solution Problem 28
Evaluate the cylindrical coordinate integrals in Exercises \(23-28\) $$ \int_{0}^{2 \pi} \int_{0}^{1} \int_{-1 / 2}^{1 / 2}\left(r^{2} \sin ^{2} \theta+z^{2}\ri
View solution Problem 29
Find the volume of the region bounded above by the paraboloid \(z=x^{2}+y^{2}\) and below by the square \(R :-1 \leq x \leq 1\) \(-1 \leq y \leq 1 .\)
View solution Problem 29
One leaf of a rose Find the area enclosed by one leaf of the rose \(r=12 \cos 3 \theta .\)
View solution