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Chapter 13: Problem 36

Find the volume of the solid trapped between the surface \(z=\cos x \cos y\) and the \(x y\) -plane, where \(-\pi \leq x \leq \pi\) \(-\pi \leq y \leq \pi\)

Short Answer

Expert verified
The volume is zero.

Step by step solution

01

Understand the Solid

The problem asks for the volume between the surface defined by \( z = \cos x \cos y \) and the \( xy \)-plane. This volume is defined over the domain \(-\pi \leq x \leq \pi\) and \(-\pi \leq y \leq \pi\).
02

Set Up the Double Integral for Volume

The volume \( V \) of the solid trapped between the surface \( z= \cos x \cos y \) and the \( xy \)-plane is given by the integral:\[V = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \cos x \cos y \, dy \, dx\]This integral calculates the sum of infinitesimally small volumes \( \cos x \cos y \, dy \, dx \) over the defined domain.
03

Integrate with Respect to y

First, integrate \( \cos x \cos y \) with respect to \( y \):\[\int_{-\pi}^{\pi} \cos x \cos y \, dy = \cos x \left[ \sin y \right]_{y=-\pi}^{y=\pi}\]Calculate the limits:\( \cos x (\sin(\pi) - \sin(-\pi)) = \cos x (0 - 0) = 0 \).
04

Integrate with Respect to x

Since the integrand's integral with respect to \( y \) is zero, the next integral:\[\int_{-\pi}^{\pi} 0 \, dx = 0\]contributes nothing to the volume. Therefore, the total volume is zero over the domain.

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Key Concepts

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

Volume of a Solid
When we talk about the volume of a solid, we're referring to the total amount of space that a three-dimensional object occupies.
This is a fundamental concept in geometry and calculus, used extensively to solve real-world problems.
For a solid lying between a surface and a plane, the volume is found by integrating the height of the solid, given by the equation of the surface, over the area of the base.
  • In this exercise, the surface is given by the function \( z = \cos x \cos y \) and the base is the rectangular region \(-\pi \leq x \leq \pi\) and \(-\pi \leq y \leq \pi\).
    The volume is computed using a double integral.
  • The double integral sums up infinitesimal volume elements \( \cos x \cos y \, dy \, dx \), with \( dy \, dx \) representing tiny areas on the base, and \( \cos x \cos y \) representing the height at each point of the base.
Ultimately, finding the volume of such solids involves applying calculus techniques to understand how these small pieces add up to the entire solid.
Surface Integral
Surface integrals extend the concept of integration to functions over surfaces.
In general, a surface integral computes the sum of a function over a particular surface, taking into account both the function's value and the shape of the surface.
In this exercise, while we're not directly performing a surface integral, the idea relates closely because the surface \( z = \cos x \cos y \) is integral to the problem.
  • Essentially, the double integral used to find the volume is a kind of surface integral where the surface is the projection onto the \( xy \)-plane.
    The value of the function \( z = \cos x \cos y \) over this surface provided the "height" of the surface above each point in the domain.
  • A classic surface integral problem might compute the flow of a vector field across a curved surface, whereas here we sum up volumes beneath the surface itself.
This highlights the versatility of integration in dealing with different dimensions, from lines and surfaces to volumes.
Integral Calculus
Integral calculus is a cornerstone in mathematics that deals with the accumulation of quantities.
Inevitably, it has a vast array of applications including calculating areas, volumes, and more.
In the case of the volume calculation for the given solid, we use a double integral, a tool from integral calculus.
  • The double integral \( \int \int \) enables us to add up infinitely many infinitesimal quantities, such as small bits of area or volume, to find a total.
  • In our example, each element \( \cos x \cos y \, dy \, dx \) is calculated by multiplying the function value at a point by a tiny area.
  • This method of integration reflects one of the most potent ways integral calculus helps solve real-world problems, by approximating and then adding up small parts to understand a whole.
Understanding integral calculus and its operations allows us to tackle a wide range of mathematical and practical challenges.

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Most popular questions from this chapter

Evaluate the given double integral by changing it to an iterated integral. $$ \begin{aligned} &\text { 16. } \iint_{S}(x+y) d A ; S \text { is the triangular region with vertices }\\\ &(0,0),(0,4), \text { and }(1,4) \end{aligned} $$.

Find the mass \(m\) and center of mass \((\bar{x}, \bar{y})\) of the lamina bounded by the given curves and with the indicated density. \(y=e^{x}, y=0, x=0, x=1 ; \delta(x, y)=2-x+y\)

In Problems \(37-39,\) evaluate each iterated integral. $$ \int_{-2}^{2} \int_{-1}^{1}\left|x^{2} y^{3}\right| d y d x $$

Evaluate the iterated integrals. $$ \int_{0}^{1} \int_{0}^{3 x} x^{2} d y d x $$

Write the given iterated integral as an iterated integral with the order of integration interchanged. Hint: Begin by sketching a region \(S\) and representing it in two ways. $$ \int_{1 / 2}^{1} \int_{x^{3}}^{x} f(x, y) d y d x $$
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