Problem 49 - Thomas Calculus

Question

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 \(2 + 2\ln 2\).

Step 1: Identify the Bounded Regions

The solid is bounded by the planes and cylinders: the planes are vertical cross-sections at \(x=1\) and \(x=2\). On the sides, the solid is bounded by the cylinders described by the equations \(y = \pm \frac{1}{x} \). This implies a region in the xy-plane from \(x=1\) to \(x=2\) and from \(y=-\frac{1}{x}\) to \(y=\frac{1}{x}\). Lastly, the solid's height is determined by \(z = x + 1\) and \(z = 0\).

Step 2: Set Up the Integral

The volume of the solid can be found using a triple integral in Cartesian coordinates. Since the solid is defined over the region \(1 \leq x \leq 2\), \(-\frac{1}{x} \leq y \leq \frac{1}{x}\), and \(0 \leq z \leq x+1\), we set up the triple integral: \[V = \int_{1}^{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} \int_{0}^{x+1} dz \, dy \, dx.\]

Step 3: Compute the Innermost Integral

First, integrate with respect to \(z\). The limits of \(z\) are from 0 to \(x+1\), so:\[\int_{0}^{x+1} dz = z \Big|_{0}^{x+1} = (x + 1) - 0 = x + 1.\]This simplifies the integral to:\[V = \int_{1}^{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} (x+1) \, dy \, dx.\]

Step 4: Compute the Middle Integral

Now, integrate with respect to \(y\). The function \(x+1\) does not depend on \(y\), so:\[\int_{-\frac{1}{x}}^{\frac{1}{x}} (x+1) \, dy = (x+1) \cdot \left( y \right) \Big|_{-\frac{1}{x}}^{\frac{1}{x}} = (x+1) \cdot \left( \frac{1}{x} - \left(-\frac{1}{x}\right) \right) = (x+1) \cdot \left( \frac{2}{x} \right).\]This becomes:\[2(x+1)/x.\]

Step 5: Compute the Outermost Integral

Now integrate the expression we obtained with respect to \(x\):\[V = \int_{1}^{2} \frac{2(x+1)}{x} \, dx = \int_{1}^{2} 2\left(\frac{x}{x} + \frac{1}{x}\right) \, dx = \int_{1}^{2} 2\left(1 + \frac{1}{x}\right) \, dx = \int_{1}^{2} 2 \, dx + \int_{1}^{2} \frac{2}{x} \, dx.\]This evaluates to:\[2[x]_{1}^{2} + 2[\ln|x|]_{1}^{2} = 2(2 - 1) + 2 (\ln 2 - \ln 1) = 2 + 2\ln 2.\]

Step 6: Final Volume

Combine the results from the outer integral evaluation to find the total volume of the solid. The volume is:\[V = 2 + 2\ln 2.\]

Key Concepts

Volume Calculation

Calculating volumes using triple integrals is a powerful technique in multivariable calculus. A triple integral extends the concept of integration to three dimensions, allowing us to find the volume of complex shapes. By integrating over a certain region, we take into account the entirety of the space the solid occupies.

For a given solid, the volume can be calculated by setting up a triple integral over the region that the solid occupies, taking care to define the limits of integration clearly. The limits for each variable define the range over which the solid extends in space. In this exercise, the volume of the solid is given as: \[V = \int_{1}^{2} \int_{-\frac{1}{x}}^{\frac{1}{x}} \int_{0}^{x+1} dz \ dy \ dx. \]

The sequence of integration here—from \(z\) to \(y\), and finally \(x\)—allows us to first consider vertical slices of the solid and then sum these slices horizontally and beyond. Each step involves solving an integral that narrows down the volume calculation increasingly from 3D to find the exact volume of the defined region.

Bounded Regions

Understanding the boundaries of a region is vital for setting up triple integrals correctly. A boundary in a volume problem defines the limits within which the entire solid is enclosed. In this problem, the boundaries are given by planes and cylinders, each providing specific limits for integration.

The solid is bounded in the \(x\)-direction by the planes \(x=1\) and \(x=2\). These limits serve as the end points for the outermost integral. In the \(y\)-direction, the boundary is formed by the equations \(y = \pm \frac{1}{x}\). This forms a curvy boundary instead of a straight line, reflecting the nature of cylinders when intersecting with the x-axis. Lastly, the \(z\)-direction is bounded by the planes \(z=0\) and \(z=x+1\), setting the ground and ceiling of the solid respectively.

Each boundary affects how we slice and integrate through the region, ensuring that the entire space described is accurately accounted for in volume calculations.

Cartesian Coordinates

The Cartesian coordinate system forms the foundation for setting up triple integrals in many cases. It uses three perpendicular axes—\(x\), \(y\), and \(z\)—to define points in 3-dimensional space, making it ideal for determining the volume of solids with defined boundaries like planes and cylinders.

In this exercise, the coordinate axes are essential for translating the mathematical inequalities given by the boundaries into clear integral limits. The choice of Cartesian coordinates is often dictated by the simplicity they bring to the problem, especially when equations of boundaries are naturally expressed in terms of \(x\), \(y\), and \(z\).

Setting up the triple integral with these coordinates allows for methodical computation, crucial for understanding both school assignments and real-world applications where the shape of a solid can be more complex.

Cylindrical Integration

While this exercise primarily uses Cartesian coordinates, integrating over regions bounded by cylinders is often a situation where cylindrical integration would shine. Cylindrical coordinates transform the usual \((x, y, z)\) system into \((r, \theta, z)\), which better describes shapes with circular symmetry, primarily around one axis.

However, in this problem, the boundaries are not complete circular cylinders but involve a combination of planar boundaries which are more straightforward in Cartesian form. Yet, the cylinder equations \(y = \pm \frac{1}{x}\) are typical of problems where one might consider cylindrical coordinates if the symmetry was more dominant.

Understanding when to use cylindrical coordinates is key. If the problem had perfect symmetric cylindrical limits, we could switch, simplifying certain equations and integrals. Recognizing shapes and situational cues for optimally choosing coordinate systems can lead to more efficient calculations in advanced contexts.