Problem 31 - Calculus Early Transcendentals

Question

The surface of an island is defined by the following functions over the region on which the function is non-negative. Find the volume of the island. $$z=25-\sqrt{x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

Answer: The volume of the island is \(\frac{625}{3}\pi \cdot 25^2\).

Step 1: Determine the region for non-negative function values

To find the range of x and y where the function is non-negative, we rewrite the inequality, \(25 - \sqrt{x^2 + y^2} \geq 0\). Solving for \(x^2 + y^2\), we get: $$x^2 + y^2 \leq 25^2$$ This represents a circle of radius 25 in the xy-plane.

Step 2: Set up the double integral

In order to find the volume under the given function, we need to take a double integral of it over the specified region. The function only depends on the distance from the origin, so it would be easier to switch to polar coordinates. So, we switch from (x,y) to (r,θ), with \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). The function will now be: $$z = 25 - \sqrt{r^{2}} = 25 - r$$ The \(x^2 + y^2 \leq 25^2\) corresponds to \(0 \leq r \leq 25\) in polar coordinates. The double integral will be $$V = \int_0^{2\pi} \int_0^{25} (25 - r)r dr d\theta$$

Step 3: Evaluate the r integral

First, we evaluate the r integral: $$\int_0^{25} (25 - r)r dr = \int_0^{25} (25r - r^2) dr = \left[\frac{25}{2}r^2 - \frac{1}{3}r^3\right]_0^{25} = \frac{625}{6} \cdot 25^2$$

Step 4: Evaluate the θ integral

Now, evaluate the θ integral: $$V = \int_0^{2\pi} \frac{625}{6} \cdot 25^2 d\theta = \frac{625\cdot 25^2}{6} [ \theta]_0^{2\pi} = \frac{625\cdot 25^2}{6} \cdot 2\pi = \frac{625}{3}\pi\cdot 25^2$$ Thus, the volume of the island is \(\frac{625}{3}\pi \cdot 25^2\).

Key Concepts

Polar Coordinates

Polar coordinates can transform our viewpoint from the familiar x and y Cartesian grid into something more suited for problems involving circles and angles. When dealing with circular or rotational symmetry in a plane, polar coordinates simplify computations, as they are based on radial distances and angles. Rather than using the x-y coordinate pair to specify a point, we use

• Radius \( r \): The distance from the origin to the point.
• Angle \( \theta \): The angle from the positive x-axis to the radius line extending to the point.

To switch from Cartesian coordinates to polar, we use

• \( x = r\cos(\theta) \)
• \( y = r\sin(\theta) \)

This transformation makes equations involving circles and angles more straightforward, an essential operation when calculating areas or volumes under curves.

Double Integration

Double integration is an extension of single-variable integration into multiple dimensions. It's a useful tool for finding the volume under a surface in three-dimensional space. In problems involving curved or flat surfaces, double integration calculates such volumes or areas by breaking the problem down into a sum of infinitesimally small areas, then adding them up. In this exercise, we're aiming to find the volume of a region defined by the function \( z = 25 - \sqrt{x^2 + y^2} \). To do this, we have transformed our coordinates to polar form, enabling us to express the volume as \[ V = \int_0^{2\pi} \int_0^{25} (25 - r)r \, dr \, d\theta \]. The double integral is evaluated in two steps: first over \( r \) (the radius) and then over \( \theta \) (the angle). This ensures we correctly account for the contribution of each tiny section of the area under the curve.

Circle in Plane

A circle in the plane is a set of all points equidistant from a central point. In this exercise, the circle is defined by the inequality \( x^2 + y^2 \leq 25^2 \). It forms a boundary for the region over which we calculate the volume. The concept of a circle is incredibly useful when working with polar coordinates, as radii and angles naturally align with circles.

• The radius of this circle is 25.
• The circle's center is at the origin (0,0).

In the context of the problem, this circular boundary confines the region of integration, dictating where the surface \( z = 25 - \sqrt{x^2 + y^2} \) lies above the xy-plane and thus where the volume is non-negative. This precise arrangement simplifies the integral setup by ensuring the limits are **zero cents worried about the matching radii** and the angle \( \theta \) spans the full circle.