Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid in the first octant under the paraboloid \(z=x^{2}+y^{2}\) and inside the cylinder \(x^{2}+y^{2}=9\) by using polar coordinates.

Step-by-Step Solution

Answer

Volume is \(\frac{81\pi}{8}\).

Step 1: Understand the Problem

We are tasked to find the volume of a solid under the paraboloid given by the equation \(z = x^2 + y^2\) and inside the cylinder \(x^2 + y^2 = 9\) in the first octant. The first octant means that \(x, y, z \geq 0\).

Step 2: Convert to Polar Coordinates

In polar coordinates, we use the relations \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). The paraboloid equation becomes \(z = r^2\), and the boundary of the cylinder becomes \(r^2 = 9\), i.e., \(r = 3\).

Step 3: Set Up the Integral

In the first octant, \(\theta\) ranges from \(0\) to \(\frac{\pi}{2}\) and \(r\) ranges from \(0\) to \(3\). The differential area element in polar coordinates is \(r \, dr \, d\theta\), so the volume integral is given by: \[V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{3} r^2 \, r \, dr \, d\theta\]This simplifies to:\[V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{3} r^3 \, dr \, d\theta\]

Step 4: Evaluate the Integral

We first integrate with respect to \(r\):\[\int_{0}^{3} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{3} = \frac{3^4}{4} - \frac{0^4}{4} = \frac{81}{4}\]Then, we integrate with respect to \(\theta\):\[V = \int_{0}^{\frac{\pi}{2}} \frac{81}{4} \, d\theta = \frac{81}{4} \Theta \Bigg|_{0}^{\frac{\pi}{2}} = \frac{81}{4} \cdot \frac{\pi}{2} = \frac{81\pi}{8}\]

Step 5: Conclude the Solution

The volume of the solid under the paraboloid and inside the cylinder in the first octant is \(\frac{81\pi}{8}\).

Key Concepts

Polar Coordinates

Polar coordinates are an alternative way to describe the location of a point in a plane. Instead of using the traditional Cartesian coordinates

• Cartesian (x, y) - uses horizontal and vertical axes.

we use distance and angle in polar coordinates. In three-dimensional space, polar coordinates become especially useful when dealing with problems that involve symmetry around an axis.

How to Use Polar Coordinates?

• Convert coordinates:
  • \(x = r \cos(\theta)\)
  • \(y = r \sin(\theta)\)
  • \(r^2 = x^2 + y^2\)
• Replace \(x\) and \(y\) in equations with their polar forms.
• Set limits for \(r\) and \(\theta\) depending on the given conditions.

Using polar coordinates simplifies equations of circular and rotational symmetry as seen in this problem involving a cylinder.

Paraboloid

A paraboloid is a three-dimensional surface characterized by parabolic sections along two dimensions and a circular section along the third. Imagine stretching a parabola along a direction perpendicular to its plane to form a three-dimensional space.

Equation of a Paraboloid

• In Cartesian coordinates, it is given by the formula \(z = x^2 + y^2\).
• This indicates that the height \(z\) of the paraboloid increases with the square of the distance from the origin in the xy-plane.

In this exercise, the paraboloid is acted as a boundary surface for the volume being calculated. It's crucial to visualize the paraboloid extending up from the center of the coordinate plane.

Cylinder

In geometry, a cylinder is a three-dimensional surface that has two parallel bases joined by a curved surface at a fixed distance from each other, forming a set of all points equidistant from a line segment known as the axis.

Equation of a Cylinder

• Expressed in Cartesian coordinates as \(x^2 + y^2 = a^2\), where \(a\) represents the radius of the circular base.
• The circle is extended vertically (or horizontally), making the cylinder's height infinite in mathematics.
• Cylindrical coordinates are ideal for dealing with circular boundaries, such as those in this problem.

This exercise required understanding cylinders in a circular boundary context defined by the equation \(x^2 + y^2 = 9\), which translates into a polar radius of \(r = 3\).

Definite Integrals

In calculus, definite integrals represent the accumulation of quantities, such as areas under curves. They provide a way to calculate the total "volume" or "area" bounded by known functions within a given region.

Properties of Definite Integrals

• Boundaries: Defined by the integration limits along each axis.
• Multivariable Integration: Extends this concept to more than one dimension, like the integral used here.
• Polar Integral: Uses the polar area element \(r \, dr \, d\theta\).

In the given exercise, understanding definite integrals allowed us to sum up small volume elements under the paraboloid and within the cylindrical boundary, producing the measured volume \(\frac{81\pi}{8}\).