Problem 17
Question
Set up the iterated integral for evaluating \(\iiint_{D} f(r, \theta, z) d z r d r d \theta\) over the given region \(D .\) \(D\) is the solid right cylinder whose base is the region in the \(x y-\) plane that lies inside the cardioid \(r=1+\cos \theta\) and outside the circle \(r=1\) and whose top lies in the plane \(z=4 .\)
Step-by-Step Solution
Verified Answer
The iterated integral is \(\int_{0}^{2\pi} \int_{1}^{1+\cos \theta} \int_{0}^{4} f(r, \theta, z)\, dz \, r \, dr \, d\theta\).
1Step 1: Understand the Region D
The solid region \(D\) is a right cylinder. The base of this cylinder is in the \(xy\)-plane between the cardioid \(r = 1 + \cos \theta\) and the circle \(r = 1\). The height of the cylinder is from \(z = 0\) to \(z = 4\).
2Step 2: Determine Limits in Polar Coordinates
Since we are dealing with polar coordinates (\(r\), \(\theta\)), set the inner radial limits. For \(r\), the region lies between the cardioid and the circle, so \(1 \leq r \leq 1 + \cos \theta\). The angle \(\theta\) runs from 0 to \(2\pi\) as these curves make a full loop around the origin.
3Step 3: Set Up the Integral
The integral, expressed in cylindrical coordinates, becomes: \[\int_{0}^{2\pi} \int_{1}^{1+\cos \theta} \int_{0}^{4} f(r, \theta, z)\, dz \, r \, dr \, d\theta\]
4Step 4: Explain the Order of Integration
The iterated integral \(\int_{0}^{2\pi} \int_{1}^{1+\cos \theta} \int_{0}^{4} f(r, \theta, z)\, dz \, r \, dr \, d\theta\) follows the order \(d z\), \(r \, d r\), \(d\theta\), since \(z\) changes from the bottom to the top of the cylinder, life within fixed radial sections described by \(r\), and \(\theta\) covers the full circle in the polar coordinate system.
Key Concepts
Iterated IntegralsPolar CoordinatesCardioid RegionLimits of Integration
Iterated Integrals
Iterated integrals are a way to evaluate integrals over multidimensional regions, like the inside of a cylinder. They break down a complex integral into simpler, one-dimensional integrals which are carried out step by step. In this case, the iterated integral is expressed in the form:
The most important aspect to note here is the nesting effect where each inner integral must be computed first. The result from the innermost integral is then used in the next, and so forth. This lays down the foundational idea of iteratively integrating in one variable before moving onto the next.
- \[\int_{0}^{2\pi} \int_{1}^{1+\cos \theta} \int_{0}^{4} f(r, \theta, z)\, dz \, r \, dr \, d\theta \]
The most important aspect to note here is the nesting effect where each inner integral must be computed first. The result from the innermost integral is then used in the next, and so forth. This lays down the foundational idea of iteratively integrating in one variable before moving onto the next.
Polar Coordinates
Polar coordinates provide a way of describing the position of a point in a plane using a radius and an angle rather than the traditional Cartesian coordinates (\(x\) and \(y\)). In mathematical and scientific problems that have symmetry around a point, like a circle or a cardioid, polar coordinates are especially useful.
In this integral, \(r\) represents the distance from the origin, while \(\theta\) is the angle from the positive \(x\)-axis.
For instance, converting Cartesian limits to polar can often leave to more concise expressions, like the transition seen when defining our limits for \(\theta\) and \(r\) in the current exercise.
In this integral, \(r\) represents the distance from the origin, while \(\theta\) is the angle from the positive \(x\)-axis.
- This allows for seamless integration when dealing with regions that form around the origin, such as circular regions.
- It simplifies the math since, in some cases, makes solving the integral a lot easier by considering radial symmetry.
For instance, converting Cartesian limits to polar can often leave to more concise expressions, like the transition seen when defining our limits for \(\theta\) and \(r\) in the current exercise.
Cardioid Region
A cardioid is a heart-shaped curve described by the equation \(r = 1 + \cos \theta\). It's a distorted circle where the distance from any point on the curve to a fixed point (the pole) remains constant.
In this problem, the region between the cardioid and a circle forms the base of a cylinder.
This helps in ensuring that all calculations regarding area, volume, or other properties confined within this boundary are accurate and logically grounded.
In this problem, the region between the cardioid and a circle forms the base of a cylinder.
- The cardioid region is defined for \(0 \leq \theta \leq 2\pi\), as it forms a complete enclosure around the origin.
- This region lies outside the circle \(r = 1\) and inside the cardioid, creating an annular shape.
This helps in ensuring that all calculations regarding area, volume, or other properties confined within this boundary are accurate and logically grounded.
Limits of Integration
Choosing correct limits of integration is vital for accurately solving an integral over a specific region. In this exercise, the integration bounds are established based on the cylinder's geometry. The cylinder's base lies in the polar coordinate plane, with \(0 \leq z \leq 4\) defining its height.
Choosing these correctly is key when converting the region from Cartesian to polar coordinates. Always verify your limits correspond to the actual regions and surfaces described. This ensures no part of the region you're attempting to integrate over is mistakenly omitted or double-counted.
- For radial limits, the integral runs from the inner circle \(r = 1\) to the outer cardioid surface \(r = 1 + \cos \theta\).
- The angle \(\theta\) varies from \(0\) to \(2\pi\), ensuring full circular coverage.
Choosing these correctly is key when converting the region from Cartesian to polar coordinates. Always verify your limits correspond to the actual regions and surfaces described. This ensures no part of the region you're attempting to integrate over is mistakenly omitted or double-counted.
Other exercises in this chapter
Problem 17
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Find the area of the region cut from the first quadrant by the curve \(r=2(2-\sin 2 \theta)^{1 / 2} .\)
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