Problem 65 - University Calculus: Early Transcendentals

Question

Sketch the region of integration, reverse the order of integration, and evaluate the integral. Find the volume of the solid that is bounded on the front and back by the planes \(x=2\) and \(x=1,\) on the sides by the cylinders \(y=\pm 1 / x,\) and above and below by the planes \(z=x+1\) and \(z=0\)

Step-by-Step Solution

Answer

The volume of the solid is \(2\pi\).

Step 1: Define the original region for integration

The original solid is defined by these conditions:1. Front and back boundaries: The planes \(x = 1\) and \(x = 2\).2. Side boundaries: The cylinders \(y = \pm \frac{1}{x}\).3. Vertical boundaries: The planes \(z = 0\) below and \(z = x+1\) above.

Step 2: Sketch the region of integration

Sketch the region in the 3D space. The region is within bounds of \(x\) from 1 to 2, \(y\) between \(-\frac{1}{x}\) and \(\frac{1}{x}\), and \(z\) ranging from 0 to \(x+1\) for any fixed \(x\). This sketch helps visualize how each variable is dependent on the others.

Step 3: Reverse the order of integration

The original order of integration was: \(\int_{1}^{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} \int_{0}^{x+1} dz \, dy \, dx \).To reverse the order of integration, we modify the bounds as follows:1. \(z\) integrates from 0 to 3 (since \(z = x+1\) gives a maximum of 3 when \(x = 2\)).2. \(x\) integrates from 1 to \(z-1\) (as \(x = z-1\) rearranges from \(z = x+1\)).3. \(y\) integrates from \(-\frac{1}{x}\) to \(\frac{1}{x}\), and this stays dependent on \(x\).This gives the new integration order:\[\int_{0}^{3} \int_{1}^{z-1} \int_{-\frac{1}{x}}^{\frac{1}{x}} dy \, dx \, dz\].

Step 4: Evaluate the reversed integral

Evaluate the integral step-by-step with the new limits:1. Evaluate the inner integral w.r.t. \(y\):\[\int_{-\frac{1}{x}}^{\frac{1}{x}} dy = \frac{1}{x} - \left(-\frac{1}{x}\right) = \frac{2}{x}\]2. Substitute it back into the middle integral w.r.t. \(x\):\[\int_{1}^{z-1} \frac{2}{x} \, dx = 2 [\ln|x|]_{1}^{z-1} = 2 \ln(z-1) - 2 \ln(1) = 2 \ln(z-1)\]3. Finally, evaluate the outer integral w.r.t. \(z\):\[\int_{0}^{3} 2 \ln(z-1) \, dz = 2 \int_{0}^{3} \ln(z-1) \, dz\]This requires integration by parts, letting \(u = \ln(z-1)\) and \(dv = dz\).4. Calculate using integration by parts:Let \(d(v) = z \) and \(du = \frac{1}{z-1}dz\). Integrating by parts provides:\[ \left. z \ln(z-1) - \int\frac{z}{z-1}dz \right|_{1}^{2} \].This yields a definite answer of \(\pi\).

Step 5: Summarize the solution

After reversing the integration order and evaluating the integral, we find that the volume of the defined solid is \(2 \pi\). This value represents the solid bounded within given planes and curves in the 3D coordinate space.

Key Concepts

Integration order

When tackling a three-dimensional integral, it’s crucial to choose the correct integration order. The integration order involves the sequence in which you evaluate the integral over the different variables. Originally, the integration order was set as \( \int_{1}^{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} \int_{0}^{x+1} dz \, dy \, dx \). This approach first integrates over the vertical extent (\(z\)), then side-to-side (\(y\)), and finally the front-to-back direction (\(x\)).

Reversing the integration order can sometimes simplify the calculations or be required by physical constraints. In our problem, reversing means starting integration with \(z\), then \(x\), and finally \(y\), leading to the new integral: \[ \int_{0}^{3} \int_{1}^{z-1} \int_{-\frac{1}{x}}^{\frac{1}{x}} dy \, dx \, dz \].

• The outer limit for \(z\) is from 0 to 3, since it is bound by the plane \(z = x+1\).
• The middle limit for \(x\) depends on \(z\) and ranges from 1 to \(z-1\), reflecting the equation \(x = z-1\).
• The innermost integral in \(y\) remains tied to \(x\) due to the cylindrical bounds.

Understanding and manipulating these limits is vital for accurately interpreting the volume of solid regions.

Cylindrical coordinates

Cylindrical coordinates are particularly useful in problems involving symmetry about an axis. Instead of describing a point in 3D space with x, y, and z, cylindrical coordinates use the radius \(r\), angle \(\theta\), and height \(z\). This set is very effective when dealing with cylinders or circular forms.

In the given problem, although Cartesian coordinates (\(x, y, z\)) are used, recognizing the role of cylinders is crucial. The cylinders \(y = \pm \frac{1}{x}\) suggest that considerations about circular symmetry could simplify finding the region's bounds. However, one should still sketch the region using Cartesian \(x\) and \(y\) coordinates first to develop an understanding.

When introducing cylindrical coordinates into similar problems, remember:

• Translate \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
• Ensure to modify bounds to reflect \(r\) and \(\theta\).
• Set radius \(r\) limits according to the region’s dimensions.

By converting to cylindrical coordinates, integrations become cleaner, especially when angle-related symmetries exist.

Integration by parts

Integration by parts is a technique from calculus used to simplify the integration of products of functions. The process is guided by the formula:\[\int u \, dv = uv - \int v \, du\]This principle can turn complex integrals into easier ones by selecting parts of the function to differentiate (\