Problem 61 - Calculus Early Transcendentals

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 above the cone \(z=r\) and below the sphere \(\rho=2,\) for \(z \geq 0,\) in the orders \(d z d r d \theta, d r d z d \theta,\) and \(d \theta d z d r\)

Step-by-Step Solution

Answer

Question: Express the given region (above the cone \(z=r\) and below the sphere \(\rho=2\) for \(z\geq 0\)) in cylindrical coordinates in the orders \(dz\,dr\,d\theta\), \(dr\,dz\,d\theta\), and \(d\theta\,dz\,dr\). Answer: 1. For the order \(dz\,dr\,d\theta\): \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,dz\,dr\,d\theta\) 2. For the order \(dr\,dz\,d\theta\): \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_z^{\sqrt{4-z^2}} f(r,\theta,z)\,dr\,dz\,d\theta\) 3. For the order \(d\theta\,dz\,dr\): \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,d\theta\,dz\,dr\)

Step 1: Sketch the region of integration

First, let's sketch the region of integration. The region is bound by the cone \(z=r\) and the sphere \(\rho=2\). Since the range for \(z\) is non-negative values only, we know that our region is in the upper half-space. Also, note that when the cone intersects the sphere, they form a circle in the \(xy\)-plane. Sketching the region, we get the following solid: [insert 3D plot of the region above the cone \(z=r\) and below the sphere \(\rho=2\) for \(z\geq0\)] Now, let's find the limits of integration for each order.

Step 2: dz dr dθ order

To express the region in the order \(dz\,dr\,d\theta\), first identify the limits of integration: - For \(z\), we know the range is between the cone \(z=r\) and the sphere \(\rho=2\). So, \(z\) starts at the cone (\(z=r\)) and goes up to the sphere. In cylindrical coordinates, the sphere is given by \(z^2 + r^2 = 4\). Solving for \(z\), we get \(z=\sqrt{4-r^2}\). Thus, the bounds for \(z\) are \(r\leq z\leq \sqrt{4-r^2}\). - For \(r\), it ranges from \(r=0\) on the \(z\)-axis up to the circle formed by the intersection of the cone and sphere. To find this point of intersection, set \(z=r\) and \(z=\sqrt{4-r^2}\) and solve for \(r\). We get \(r^2=4-r^2\), which gives \(r=\sqrt{2}\), so the bounds for \(r\) are \(0\leq r\leq \sqrt{2}\). - Lastly, for \(\theta\), it covers the entire circular cross-section, so the bounds are \(0\leq \theta \leq 2\pi\). So, the iterated integral in the order \(dz\,dr\,d\theta\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,dz\,dr\,d\theta\)

Step 3: dr dz dθ order

Now, we'll find the limits of integration for the order \(dr\,dz\,d\theta\): - For \(r\), the range is from the cone \(z=r\) to the outer radius of the region based on the \(z\) value. To find the outer radius, we use the equation of the sphere, \(z^2 + r^2 = 4\). Solving for \(r\), we get \(r=\sqrt{4-z^2}\). Thus, the bounds for \(r\) are \(z\leq r\leq \sqrt{4-z^2}\). - For \(z\), it ranges from the vertex of the cone (\(z=0\)) up to the height where the cone and sphere intersect. From the previous step, we calculated this as \(z=\sqrt{2}\). Thus, the bounds for \(z\) are \(0\leq z\leq \sqrt{2}\). - For \(\theta\), it covers the entire circular cross-section as before, so the bounds are \(0\leq \theta \leq 2\pi\). So, the iterated integral in the order \(dr\,dz\,d\theta\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_z^{\sqrt{4-z^2}} f(r,\theta,z)\,dr\,dz\,d\theta\)

Step 4: dθ dz dr order

Lastly, we'll find the limits of integration for the order \(d\theta\,dz\,dr\): - For \(\theta\), it covers the entire circular cross-section, so the bounds are \(0\leq \theta \leq 2\pi\). - For \(z\), given \(r\), the range is from the cone (\(z=r\)) to the sphere, which was found previously as \(z=\sqrt{4-r^2}\), so the bounds for \(z\) are \(r\leq z\leq \sqrt{4-r^2}\). - For \(r\), it ranges from \(0\) at the \(z\)-axis to the maximum radius value, which we computed as \(r=\sqrt{2}\) in the previous steps. Thus, the bounds for \(r\) are \(0\leq r\leq \sqrt{2}\). So, the iterated integral in the order \(d\theta\,dz\,dr\) is: \(\int_0^{2\pi}\int_0^{\sqrt{2}}\int_r^{\sqrt{4-r^2}} f(r,\theta,z)\,d\theta\,dz\,dr\)

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system where each point in space is determined by the radial distance from the origin \(r\), the angle \(\theta\) from the positive x-axis, and the height \(z\) above the xy-plane. This system is particularly useful in situations involving cylinders or cylindrical symmetry, as it simplifies integrals and equations.
One of the challenges when solving integration problems in three dimensions is converting the region of integration into cylindrical coordinates. With cylindrical coordinates, we replace traditional Cartesian coordinates \((x, y, z)\) with \((r, \theta, z)\). Here are key transformations:

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

This approach simplifies the mathematical expressions for many solids, such as cones and spheres, and allows us to represent surfaces and volumes more intuitively.

Iterated Integrals

Iterated integrals are a method used in multivariate calculus to compute multiple integrals, especially when evaluating definite integrals over a region. The process involves carrying out multiple integrations sequentially, one integral at a time.
Each integral in an iterated integral will have its differential, such as \( dr \), \( d\theta \), or \( dz \), and corresponding limits. Different orders of integration imply that integrals must be computed as inner and outer integrals according to the chosen sequence.
For example, in the integration order \(dz\,dr\,d\theta\), \(z\) is the innermost, \(r\) is the middle, and \(\theta\) is the outermost integration variable. The iterated nature requires careful consideration of the bounds of integration for each coordinate, which often vary depending on the nested variables.

Integration Limits

Integration limits define the bounds over which the integral is calculated. In the case of a volume integral using cylindrical coordinates, limits must be carefully described for each of the cylindrical coordinates: \(z\), \(r\), and \(\theta\).
Setting these limits appropriately is crucial for accurately representing the region of integration. The limits can vary significantly depending on the specific order of integrating, and they generally stem from geometric constraints of the problem, like the intersection of a cone and a sphere in our example.

• For \(z\), the limits are determined by the cone and the sphere equations.
• For \(r\), they depend on where the solid starts from and ends radially.
• For \(\theta\), they often encapsulate a full circular rotation (i.e., \(0 \leq \theta \leq 2\pi\)).

Adjusting the order of integration can substantially simplify the integration process, even for complex regions. Recognizing and defining clear limits is vital for executing iterated integrals successfully.