Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Volume of the solid bounded by the cylinders \(x^{2}=y\) and \(z^{2}=y\) and the plane \(y=1\)

Step-by-Step Solution

Answer

The volume is 2 cubic units.

Step 1: Interpret the given equations

The problem involves two cylinders given by the equations: \(x^2 = y\) and \(z^2 = y\), and a plane \(y = 1\). The task is to find the volume of the region enclosed by these cylinders and capped at \(y = 1\).

Step 2: Determine limits of integration for y

The range for \(y\) is determined by its minimum and maximum values based on the plane that bounds it at \(y=1\). Since both cylinders are symmetrical and intersect from \(y = 0\) to \(y = 1\), the limits for \(y\) are 0 to 1.

Step 3: Identify limits and expressions for x and z

For a given \(y\), \(x\) can vary between the cylinders from \(-\sqrt{y}\) to \(\sqrt{y}\), and similarly, \(z\) can vary between \(-\sqrt{y}\) to \(\sqrt{y}\).

Step 4: Set up the triple integral for volume

The volume \(V\) of the region can be expressed as a triple integral: \[ V = \int_{y=0}^{1} \int_{x=-\sqrt{y}}^{\sqrt{y}} \int_{z=-\sqrt{y}}^{\sqrt{y}} dz \, dx \, dy \]

Step 5: Compute the inner integral with respect to z

Compute the integral with respect to \(z\): \[ \int_{z=-\sqrt{y}}^{\sqrt{y}} dz = \Bigg[ z \Bigg]_{-\sqrt{y}}^{\sqrt{y}} = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y} \]

Step 6: Integrate with respect to x

Substitute back into the integral with respect to \(x\):\[ \int_{x=-\sqrt{y}}^{\sqrt{y}} 2 \sqrt{y} \, dx = 2 \sqrt{y} \big[ x \big]_{-\sqrt{y}}^{\sqrt{y}} = 2\sqrt{y} (\sqrt{y} - (-\sqrt{y})) = 4y \]

Step 7: Final integration with respect to y

Now integrate with respect to \(y\):\[ V = \int_{0}^{1} 4y \, dy = 4 \left[ \frac{y^2}{2} \right]_{0}^{1} = 4 \times \frac{1}{2} = 2 \]

Step 8: Interpret the result

The final result indicates that the volume of the solid bounded by the given cylinders and the plane is 2 cubic units.

Key Concepts

Triple Integral

A triple integral is used to find volumes of regions in three-dimensional space. When we set up a triple integral, we are essentially stacking tiny volume elements on top of each other to fill up the entire region. The triple integral is made up of three single integrals. Each one integrates over one specific dimension (e.g., x, y, or z). This process allows us to accumulate all the volume elements inside a given boundary.

For example, if we have a solid bounded by two functions or surfaces, a triple integral can help calculate its volume. In our specific exercise, we used a triple integral to find the volume enclosed by y-bound cylinders and a plane. The important part of working with triple integrals is understanding how each function or boundary affects the limits of integration in each dimension.

Volume Calculation

Volume calculation using triple integrals is an extension of finding areas using double integrals. Here, you calculate the solid's volume within specific boundaries in three-dimensional space. For any volume calculation using triple integrals, understanding the order of integration is crucial. It defines how the solid is stacked and summed.

In this exercise, the volume of the solid is calculated by setting up a triple integral over the variables x, y, and z. We start by integrating with respect to z, then x, and finally y, embodying the volume through precise limits provided by the bounding surfaces, like the cylinders and the plane in our example. The integration procedure's primary aim is to go through each consecutive integral, tackling one variable at a time, leading us to the solid's complete volume.

Cylindrical Boundaries

Cylindrical boundaries are surfaces that bound the volume of a solid in a circular or parabolic manner. In the given problem, the cylinders are defined by equations of the form:

• \(x^2 = y\) which implies a parabolic cylinder opening along the y-axis with variable x.
• \(z^2 = y\) another parabolic cylinder, this time involving the variable z.

These equations represent boundaries in a 3D space where y dictates how far this shape extends.

To find the volume between them, it is crucial to determine how these boundaries guide the limits of integration as each of the variables varies within calculated bounds. Visualizing these boundaries can immensely help in understanding how the limits of x and z change with each value of y.

Integration Limits

Integration limits define the start and stop points for each integral in the triple integral setup. They encapsulate the region over which you want to integrate. Deciding the limits correctly ensures the volume calculation includes the entire region desired, not more or less.

In this exercise, the limits were determined by the geometry of the problem:

• For y, the limits are from 0 to 1, established by the plane \(y = 1\) intersecting the cylinders.
• For x, the range between \(-\sqrt{y}\) to \(\sqrt{y}\) follows the equation \(x^2 = y\), showing x's extension within the y-bound cylinder.
• Similarly, z follows \(-\sqrt{y}\) to \(\sqrt{y}\) because of the boundary \(z^2 = y\).

Properly setting integration limits in problems with cylindrical boundaries ensures the accurate calculus application, leading to the correct volume outcomes.