Question

Changing order of integration If possible, write iterated integrals in cylindrical coordinates for the following regions in the specified orders. Sketch the region of integration. The solid outside the cylinder \(r=1\) and inside the sphere \(\rho=5\) for \(z \geq 0,\) in the orders \(d z d r d \theta, d r d z d \theta,\) and \(d \theta d z d r\)

Short answer

The region is the solid that lies outside the cylinder with radius 1 and inside the sphere with radius 5, specifically for z ≥ 0. The region, expressed in cylindrical coordinates, has the following bounds: - \(0 \leq \theta \leq 2\pi\) - \(0 \leq z \leq 5\) - \(1 \leq r \leq \sqrt{25 - z^2}\) The iterated integrals for the given orders of integration are: 1. \(dz dr d\theta\) order: \(\int_{0}^{2\pi} \int_{1}^{\sqrt{25 - z^2}} \int_{0}^{5} dz dr d\theta\) 2. \(dr dz d\theta\) order: \(\int_{0}^{2\pi} \int_{0}^{5} \int_{1}^{\sqrt{25 - z^2}} dr dz d\theta\) 3. \(d\theta dz dr\) order: \(\int_{1}^{\sqrt{25 - z^2}} \int_{0}^{5} \int_{0}^{2\pi} d\theta dz dr\)

Step-by-step solution

Sketch the Region of Integration

We are given a solid region outside the cylinder \(r=1\) and inside the sphere \(\rho=5\) with \(z≥0\). To sketch the region, we can first draw the cylinder \(r=1\) and the sphere \(\rho=5\). We see that the lower and upper limits of r and z will be important for setting up the iterated integrals. In the sketch, we can visualize that the lower limit of r changes when it crosses the cylinder and goes from surface of the cylinder \(r=1\) to the intersection of the cylinder and sphere.

Set Up Integration Bounds

Before writing the iterated integrals, we need to determine the boundaries for r, \(\theta\), and z in this specific region. For \(\theta\): Since we are considering a solid that is outside a cylinder and inside a sphere and full rotation around z-axis, the bounds will be: $$0 \leq \theta \leq 2\pi$$ For z: We are given the condition \(z \geq 0\). Therefore, the lower bound for z will be 0. The upper bound for z will be the maximum height of the sphere: $$0 \leq z \leq 5$$ For r: Since we are outside the cylinder of radius 1, r will always start at r = 1. The upper bound of r is determined by where the sphere intersects the plane of a specific angle \(\theta\). We can find that using the coordinate conversion: \(x^2 + y^2 + z^2 = \rho^2 = 5^2 = 25\) \(r^2 + z^2 = 25\) \(r = \sqrt{25 - z^2}\) $$1 \leq r \leq \sqrt{25 - z^2}$$ Now, we can write the iterated integrals in the given orders.

Iterated Integral in \(dz dr d\theta\) Order

Integral in the order \(dz dr d\theta\): $$\int_{0}^{2\pi} \int_{1}^{\sqrt{25 - z^2}} \int_{0}^{5} dz dr d\theta$$

Iterated Integral in \(dr dz d\theta\) Order

Integral in the order \(dr dz d\theta\): $$\int_{0}^{2\pi} \int_{0}^{5} \int_{1}^{\sqrt{25 - z^2}} dr dz d\theta$$

Iterated Integral in \(d\theta dz dr\) Order

Integral in the order \(d\theta dz dr\): $$\int_{1}^{\sqrt{25 - z^2}} \int_{0}^{5} \int_{0}^{2\pi} d\theta dz dr$$ So, these are the iterated integrals for the given solid region in cylindrical coordinates for the specified orders of integration.

Key concepts

Iterated Integrals

Iterated integrals are used to compute multiple integrals of functions over a specified region. In the context of cylindrical coordinates, they help us integrate over a three-dimensional space by breaking down complex integrals into simpler, single-variable integrals. This means we tackle the problem one variable at a time, making it more manageable.
For our exercise, we are looking at a region outside a cylinder and inside a sphere, which fits perfectly for cylindrical coordinates. The iterated integrals are expressed in different orders, such as \(d z d r d \theta\), and are written to cover the height, radius, and full rotation around the z-axis. Each integral accounts for one of these dimensions, showing us how we can traverse through the entire region.

Order of Integration

Order of integration refers to the sequence in which we integrate the variables in an iterated integral. Depending on the problem, one might change this order to simplify the calculations or to fit the integration to the described geometry more appropriately.
In our problem, changing the order from \(d z d r d \theta\) to \(d r d z d \theta\) or \(d \theta d z d r\) does not change the final result of the integral but affects the intermediate steps. Selecting an effective order can help simplify boundary calculations and integration, especially when dealing with complex regions. Each integration order gives us insight into how we approach integrating different parts of the solid region.

Integration Bounds

Integration bounds are the limits within which each variable is integrated. They define the specific area or volume over which the integration takes place. In cylindrical coordinates, these bounds are often determined by the geometry of the problem:

• For the azimuthal angle \(\theta\), it typically varies from \(0\) to \(2\pi\) for a full rotation around the z-axis.
• For the vertical coordinate \(z\), the bounds are defined from the bottom of the region up to the top, such as \(z \geq 0\).
• For the radial coordinate \(r\), the bounds are defined from the inner to the outer edge, considering surfaces like cylinders or spheres, as seen in \(1 \leq r \leq \sqrt{25 - z^2}\).

The careful determination of these bounds is crucial to accurately representing the volume or area described in the problem.

Cylindrical to Cartesian Conversion

Converting between cylindrical and Cartesian coordinates is often necessary to relate the geometry of the region or the function itself more directly to the problem at hand.
In cylindrical coordinates, we use \(r\), \(\theta\), and \(z\), which correspond to radius, angle, and height, respectively. The conversion formulas are:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)
• \(z = z\)

These conversions help us to translate problems from one coordinate system to another, facilitating the use of integrals or visualizing the region, like a cylinder or sphere, more effectively in the space. By understanding both systems, we can tackle more complex three-dimensional problems with confidence.