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Chapter 18: Problem 10

Show that a double integral \(\iint_{R} f(x, y) d A\) is a special case of a surface integral.

Short Answer

Expert verified
A double integral is a surface integral over the surface \(z = f(x, y)\).

Step by step solution

01

Conceptual Understanding

First, let's understand the concepts of both a double integral and a surface integral. A double integral \(\iint_{R} f(x, y) dA\) computes the volume under the surface \(z = f(x, y)\) over a region \(R\) in the \(xy\)-plane. A surface integral \(\iint_{S} g(x, y, z) dS\) integrates over a surface in three-dimensional space.
02

Relate Double Integral to Surface Integral

To show that a double integral is a special case of a surface integral, consider a specific surface \((x, y, f(x, y)))\) where the function \(f(x, y)\) projects vertically from the plane \(z = 0\). Here, the surface \(S\) is the graph of \(z = f(x, y)\).
03

Set the Integrals Equivalent

When the surface \(S\) is described as \(z = f(x, y)\), the surface area element \(dS\) collapses to \(dA\) on the \(xy\)-plane, because the \`projection\` of this surface onto the \(xy\)-plane is the region \(R\). Thus, \(\iint_{R} f(x, y) dA = \iint_{S} 1 \, dS\) with \(z = f(x, y)\), showing the double integral as a surface integral.
04

Conclusion

Therefore, by considering the function \(g(x, y, z) = 1\) and reducing \(dS\) to \(dA\) on the plane, the double integral of \(f(x, y)\) over \(R\) can be viewed as a special case of a surface integral over the surface \(z = f(x, y)\).

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Key Concepts

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

Surface Integral
A surface integral is a concept that extends the idea of integration from two-dimensional regions to surfaces in three-dimensional space. Imagine you have a curved surface, like a hillside. To compute a surface integral, you sum up values over every tiny patch of the surface.
These integrals are expressed mathematically as \( \iint_{S} g(x, y, z) \, dS \), where \( S \) represents the surface, and \( g(x, y, z) \) is the function you integrate.
  • The function \( g(x, y, z) \) can represent various quantities, such as mass density or electric charge.
  • The element \( dS \) corresponds to a tiny area of the surface.
In the context of the given problem, a critical step is to see how surface integrals relate to double integrals. A double integral can be seen as a surface integral where the surface is flat and lies in a specific plane.
Volume Under a Surface
The volume under a surface is beautifully calculated using a double integral. If you think about a surface above a flat region, the double integral helps compute the volume trapped between the surface and the region below in the xy-plane.
Mathematically, this is represented by the double integral \( \iint_{R} f(x, y) \, dA \), where
  • \( R \) is the region in the xy-plane you are integrating over.
  • \( f(x, y) \) is the function that defines the surface above this region.
Visualizing it can be like picturing a canopy floating over a flat land. The region under this canopy is equivalent to the integral's output. Double integrals are powerful in many applied mathematics fields, often used for finding volumes of more complex shapes.
Region in the XY-Plane
The region in the xy-plane is the foundational base over which many integrations depend, particularly in double integrals. It is essentially a two-dimensional section that sets the boundaries for where the surface "sits" or "projects".
This region \( R \) defines where the calculations occur and determines the limits for double integrals.
  • The region in the xy-plane can be any shape - like rectangles, circles, or more complex forms.
  • By defining this region, you establish the scope and scale for finding areas, volumes, and more.
So, when considering a surface integral transformed into a double integral, always focus on the region \( R \) in the xy-plane first. The boundaries of integration for the double integral directly stem from this region, making it crucial for understanding the integration process.

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Most popular questions from this chapter

Use Green's theorem to evaluate the line integral. \(\oint_{c} x y d x+(y+x) d y\) \(C\) is the circle \(x^{2}+y^{2}=1\)

Evaluate $$\int_{C} x d x+(x+y) d y+(x+y+z) d z$$ for the given curve \(C\) from \(A(0,0,0)\) to \(B(2,4,8)\). \(C\) consists of the line segment from \(A\) to \(B\).

If \(\int_{C} M(x, y, z) d x+N(x, y, z) d y+P(x, y, z) d z\) is inde- pendent of path and \(M, N,\) and \(P\) have continuous first partial derivatives, prove that $$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}, \quad \frac{\partial M}{\partial z}=\frac{\partial P}{\partial x}, \quad \frac{\partial N}{\partial z}=\frac{\partial P}{\partial y} $$.

If \(\mathbf{F}(x, y, z)=2 x e^{2 y} \mathbf{i}+2\left(x^{2} e^{2 y}+y \cot z\right) \mathbf{j}-y^{2} \csc ^{2} z \mathbf{k}\). prove that \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is independent of path, and find a potential function \(f\) for \(\mathbf{F}\).

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