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Chapter 16: Problem 78

In Problems \(78-80\), give an example of: An iterated double integral, with limits of integration. giving the volume of a cylinder standing vertically with a circular base in the \(x y\) -plane.

Short Answer

Expert verified
Volume is given by \( h\pi R^2 \) using \( \int_{0}^{2\pi} \int_{0}^{R} h \, r \, dr \, d\theta \).

Step by step solution

01

Understanding the Problem

We are tasked with finding the volume of a vertically standing cylinder with a circular base in the xy-plane. An iterated double integral will help us find the volume.
02

Defining the Cylinder

Assume the circular base has radius \( r \) and is centered at the origin \((0,0)\), while the height of the cylinder is \( h \). This gives us a cylinder with the equation \( x^2 + y^2 \leq r^2 \).
03

Setting the Limits of Integration

The limits for \( x \) and \( y \) derive from the circle's equation. For polar coordinates: \( x = r\cos(\theta) \), \( y = r\sin(\theta) \), with \( 0 \leq \theta \leq 2\pi \) and \( 0 \leq r \leq R \).
04

Formulating the Double Integral

Convert the integral to polar coordinates to ease calculation:\[ \int_{0}^{2\pi} \int_{0}^{R} h \, r \, dr \, d\theta\] This integral calculates the volume by multiplying the height \( h \) by the area of the base.
05

Solving the Integral

First, evaluate the inner integral with respect to \( r \):\[ \int_{0}^{R} h \, r \, dr = h \left[ \frac{r^2}{2} \right]_{0}^{R} = h \frac{R^2}{2}\]Now, integrate with respect to \( \theta \):\[ \int_{0}^{2\pi} h \frac{R^2}{2} \, d\theta = h \frac{R^2}{2} \left[ \theta \right]_{0}^{2\pi} = h \frac{R^2}{2} \times 2\pi = h\pi R^2\]
06

Conclusion

The result \( h\pi R^2 \) confirms the formula for the volume of a cylinder, showing the double integral correctly represents the cylinder's volume.

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Key Concepts

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

Volume of a Cylinder
Understanding the volume of a cylinder is fundamental when working with double integrals. A cylinder typically has a circular base and a specific height. Imagine a can of soda standing upright; its round top and bottom form the bases. To find the volume, you multiply the area of the circular base by the height. The area of a circle is calculated using the formula \(\pi R^2\), where \(R\) is the radius of the circle. Therefore, the volume \(V\) of a cylinder can be expressed as:
  • \(V = \pi R^2 h\), where \(h\) is the height of the cylinder.
In calculus, this formula can be derived using double integrals, especially when determining the volume in cases where the base is not perfectly flat or is defined by a more complex equation.
Polar Coordinates
Polar coordinates are an alternative to Cartesian coordinates, which can simplify integration in cases involving circular shapes. They express a point in terms of its distance from a central point and its angle from a particular direction (usually the positive x-axis). For a point \(P\) in the plane, the polar coordinates are represented by \( (r, \theta)\).
  • \(r\) is the radius or distance from the origin.
  • \(\theta\) is the angle between the radius vector and the positive x-axis.
In the context of a cylinder with a circular base centered at the origin in the xy-plane:
  • The conversion from Cartesian coordinates to polar coordinates is given by \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
Using polar coordinates can simplify the integration process when dealing with circular boundaries, as they naturally accommodate the geometry of the circle.
Limits of Integration
The limits of integration define the boundaries over which you integrate. In a double integral, you integrate over a region in the plane. For a cylinder with a circular base, the limits are determined by the radius of the base and the geometry of that base in the xy-plane.When converting to polar coordinates, the limits are defined as follows:
  • The variable \(\theta\) ranges from \(0\) to \(2\pi\), covering a full circle around the origin.
  • The variable \(r\), representing radius, ranges from \(0\) to \(R\), where \(R\) is the cylinder's base radius.
These limits ensure that the double integral covers the entire circular base during the integration process. This completeness is necessary for accurately calculating the full volume of the cylinder.
Iterated Integrals
An iterated integral is a method of evaluating multiple integrals in succession, simplifying the computation of areas and volumes in a systematic manner. In the case of a double integral, you perform integration one variable at a time, using nesting.The steps for iterating the integral over a circular base are:
  • First, integrate with respect to the radius \(r\), using the formula \[\int_{0}^{R} h \, r \, dr\]\ to calculate the contribution to the volume from the radial component.
  • Next, integrate with respect to the angle \(\theta\): \[\int_{0}^{2\pi} \]\, to accumulate the contributions from all points around the circle.
The \(r\) term in the first integral accounts for the circular shape area element in polar coordinates. Completing this process gives the volume of the cylinder and verifies the formula \(h\pi R^2\), showing how calculus captures the volume.

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Most popular questions from this chapter

Explain what is wrong with the statement. \(\int_{1}^{2} \int_{0}^{\sqrt{4-x^{2}}} 1 d y d x=\int_{0}^{\pi / 2} \int_{1}^{2} r d r d \theta\)

Let \(p(x, y)\) be a joint density function for \(x\) and \(y .\) Are the following statements true or false? \(\int_{a}^{b} p(x, y) d x\) is the probability that \(a \leq x \leq b\)

Let \(p\) be the joint density function such that \(p(x, y)=x y\) in \(R,\) the rectangle \(0 \leq x \leq 2,0 \leq y \leq 1\) and \(p(x, y)=0\) outside \(R .\) Find the fraction of the population satisfying the given constraints. $$x+y \leq 1$$

Are the statements true or false? Give reasons for your answer. The iterated integrals \(\int_{-1}^{1} \int_{0}^{1} \int_{0}^{1-x^{2}} f d z d y d x\) and \(\int_{0}^{1} \int_{0}^{1} \int_{-\sqrt{1-z}}^{\sqrt{1-z}} f d x d y d z\) are equal.

Give an example of: An integral in spherical coordinates that gives the volume of a hemisphere.
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