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An integral can be interpreted as giving the signed area over an interval; a double integral can be interpreted as giving the signed ________ over a region.

Short Answer

Expert verified
A double integral gives the signed volume over a region.

Step by step solution

01

Understanding the Problem

The problem asks us to find the concept related to a double integral, similar to how an integral gives the signed area over an interval.
02

Identifying Key Concepts

For a single integral, we often interpret it as finding the area under a curve between two points. A double integral extends this concept to functions of two variables.
03

Introducing Double Integrals

A double integral is used for functions of two variables, typically over a region in the xy-plane. It's often interpreted as a way to accumulate a quantity over an area.
04

Conceptual Interpretation of Double Integrals

Similar to how an integral provides an area, a double integral gives the signed volume of the region above the xy-plane below the surface defined by a function.

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Key Concepts

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

Area Under a Curve
When we talk about the 'area under a curve,' we usually refer to calculating the total space between the curve and the x-axis, across a specified interval. This is achieved using definite integrals. You may wonder why this is important. Measuring the area under a curve can help us understand how one variable accumulates in relation to another. For example, if you're looking at a speed-time graph, the area under the curve would represent the total distance traveled.

  • We use definite integrals \( \int_a^b f(x) \, dx \) to find this area.
  • The function \( f(x) \) represents the curve, and \( a \) and \( b \) are the endpoints of the interval.
  • The result is a numerical value that represents the total accumulated area.


Understanding this concept is crucial because it forms the foundation for more complex ideas, such as double integrals, where we explore the accumulation of area in two dimensions.
Signed Volume
The concept of 'signed volume' comes into play when we move from single integrals to double integrals. Imagine you are given a surface plot, and you want to find out not just the area it covers, but the three-dimensional volume it "occupies" between two planes. This is where double integrals shine.

  • A double integral takes a function of two variables, \( f(x, y) \), and evaluates it over a region in the \( xy \)-plane.
  • The process is similar to finding the area under a curve, but in this case, we find the "signed volume" under the surface \( z = f(x, y) \).
  • The result reflects how much "space" the surface encompasses with respect to the xy-plane.


The term "signed" indicates that the volume can be positive or negative, depending on whether the surface is above or below the xy-plane in your coordinate system.
Functions of Two Variables
A vital concept in using double integrals is understanding functions of two variables. These functions, expressed as \( f(x, y) \), involve two independent variables that affect a single dependent variable. It is like working on a map where every point \( (x, y) \) gives you a different altitude or third value.

  • These functions are often visualized as surfaces in three dimensions.
  • For example, think of a landscape with hills and valleys; each \( (x, y) \) point gives a different height or depth, which is \( f(x, y) \).
  • Double integrals over these surfaces allow us to "accumulate" information, capturing complex relationships in a more complete way than single-variable functions.


Understanding these functions gives us the groundwork to apply double integrals meaningfully, extending basic integration concepts to analyze and interpret multi-variable scenarios effectively.

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Most popular questions from this chapter

In Exercises \(15-16,\) special double integrals are presented that are especially well suited for evaluation in polar coordinates. Consider \(\iint_{f} e^{-\left(x^{2}+y^{2}\right)} d A .\) (a) Why is this integral difficult to evaluate in rectangular coordinates, regardless of the region \(R ?\) (b) Let \(R\) be the region bounded by the circle of radius \(a\) centered at the origin. Evaluate the double integral using polar coordinates. (c) Take the limit of your answer from \((b),\) as \(a \rightarrow \infty\). What does this imply about the volume under the surface of \(e^{-\left(x^{2}+y^{2}\right)}\) over the entire \(x-y\) plane?

Find the area of the given surface over the region \(R\). \(f(x, y)=\frac{2}{3} x^{3 / 2}+2 y^{3 / 2}\) over \(R,\) the rectangle with opposite corners (0,0) and (1,1).

Find the area of the given surface over the region \(R\). \(f(x, y)=3 x-7 y+2 ; R\) is the rectangle with opposite corners (-1,0) and (1,3).

Find the center of mass of the lamina described by the region \(R\) in the plane and its density function \(\delta(x, y)\) Note: these are the same lamina as in Exercises \(11-18\). \(R\) is the circle sector bounded by \(x^{2}+y^{2}=25\) in the first quadrant; \(\delta(x, y)=\left(\sqrt{x^{2}+y^{2}}+1\right) \mathrm{kg} / \mathrm{m}^{2}\)

In Exercises \(15-16,\) special double integrals are presented that are especially well suited for evaluation in polar coordinates. The surface of a right circular cone with height \(h\) and base radius \(a\) can be described by the equation \(f(x, y)=\) \(h-h \sqrt{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}},\) where the tip of the cone lies at \((0,0, h)\) and the circular base lies in the \(x\) -y plane, centered at the origin. Confirm that the volume of a right circular cone with height \(h\) and base radius \(a\) is \(V=\frac{1}{3} \pi a^{2} h\) by evaluating \(\iint_{R} f(x, y) d A\) in polar coordinates.

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