Question

In Problems 7-14, use cylindrical coordinates to find the indicated quantity. Volume of the solid bounded by the paraboloid \(z=x^{2}+y^{2}\) and the plane \(z=4\)

Short answer

The volume of the solid is \(8\pi\).

Step-by-step solution

Understand the Problem

We need to find the volume of the region bounded by the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 4 \). In cylindrical coordinates, \( x^2 + y^2 \) can be replaced by \( r^2 \), which simplifies our paraboloid equation to \( z = r^2 \).

Convert to Cylindrical Coordinates

In cylindrical coordinates, the equations become \( z = r^2 \) and \( z = 4 \). Thus the region is bounded by \( z \) from \( r^2 \) to 4 and \( r \) from 0 to the boundary where \( r^2 = 4 \) (since \( z = r^2 \) cannot exceed \( z = 4 \)). So the upper limit for \( r \) is 2.

Set up the Integral

The volume \( V \) is given by the integral in cylindrical coordinates: \[V = \int_{0}^{2} \int_{0}^{2\pi} \int_{r^2}^{4} r \, dz \, d\theta \, dr\] Here, \( r \, dz \, d\theta \, dr \) represents the volume element in cylindrical coordinates.

Integrate Over \( z \)

Perform the integration over \( z \):\[\int_{r^2}^{4} 1 \, dz = [z]_{r^2}^{4} = 4 - r^2\] So the outer integral becomes:\[\int_{0}^{2} \int_{0}^{2\pi} (4 - r^2)r \, d\theta \, dr\]

Integrate Over \( \theta \)

Since the integrand does not depend on \( \theta \), the integral over \( \theta \) simplifies to:\[\int_{0}^{2\pi} d\theta = 2\pi\] Substitute this into the expression:\[2\pi\int_{0}^{2} (4r - r^3) \, dr\]

Integrate Over \( r \)

Now perform the integration over \( r \):\[\int_{0}^{2} (4r - r^3) \, dr = \left[ 2r^2 - \frac{r^4}{4} \right]_{0}^{2}\] Calculating the bounds:\[= \left( 2(2)^2 - \frac{(2)^4}{4} \right) - (2(0)^2 - \frac{(0)^4}{4}) = 8 - 4 = 4\]

Calculate Final Volume

Multiply by the factor from the \(\theta\) integration:\[V = 2\pi \times 4 = 8\pi\]Thus, the volume of the solid is \(8\pi\).

Key concepts

Volume of Solids

Understanding the volume of solids in mathematical terms is vital for tackling various engineering and physics problems. Calculating the volume is about measuring the space a 3-dimensional shape occupies.
In our exercise, we focus on a solid bounded by a paraboloid and a plane. This region is a common example in calculus because it introduces students to interesting and practical applications of integration.
Volumes can indeed be determined using multiple techniques, but when it comes to more complex shapes like paraboloids, integral calculus proves indispensable.

Multiple Integrals

Multiple integrals extend the concept of integration from calculating areas under curves in two dimensions to dealing with volumes in three dimensions. The original problem involves integrating in cylindrical coordinates, making it simpler to handle the region bounded by a paraboloid.

• Double integrals: These are used for calculating volume when one needs to integrate a function over a two-dimensional region.
• Triple integrals: Although not directly used in this particular problem, these could extend to higher dimensions.

For the given solid, we perform an integration over the volume by treating it as step-by-step layers from the bottom to top of the solid. By first integrating over the height (z-direction), then angle (theta-direction), and finally radius (r-direction), the volume computation is both structured and intuitive.

Paraboloid

A paraboloid in geometry is a quadric surface that can be considered a 3D analog of a parabola. In this exercise, the given paraboloid is defined by the equation \(z = x^2 + y^2\). This simple form is ideal for introducing calculus students to solids of revolution.
Using cylindrical coordinates helps simplify calculations because they align naturally with the curves of cylindrical shapes. The problem shows how the region bounded by this surface and a plane can be addressed by converting to cylindrical coordinates and performing integration over these curved surfaces.
These surfaces are prevalent in nature and technology, appearing in telescopic mirrors, satellite dishes, and more, making this analysis both educational and practical.

Integration Techniques

Mastering integration techniques is pivotal for efficiently solving calculus problems like finding the volume of complex solids. In our exercise:

• Conversion to cylindrical coordinates transforms the problem into one that is more easily managed, aligning with the symmetry of the paraboloid.
• The integration itself involves a step-by-step approach: first over the height \(z\), simplifying the integral with \(d\theta\), and finally using \(dr\) to encompass the radial component.
• The use of bounds based on physical limits of the solid, specifically how \(r\) and \(z\) interact, underlines the importance of understanding the geometry of the solid.

This problem underscores the value of strategic thinking in calculus, where identifying the optimal method for integrating a function can greatly ease the computational process.