Question

Island problems The surface of an island is defined by the following functions over the region on which the function is nonnegative. Find the volume of the island. $$z=e^{-\left(x^{2}+y^{2}\right) / 8}-e^{-2}$$

Short answer

Answer: The volume of the island is approximately 15.3623 cubic units.

Step-by-step solution

Find the region of nonnegative values for the function

To determine the region where the function $$z=e^{-\left(x^{2}+y^{2}\right) / 8}-e^{-2}$$ takes nonnegative values, we need to find where $$z \geq 0$$. To do that, consider the inequality $$e^{-\left(x^{2}+y^{2}\right) / 8} \geq e^{-2}$$. Now we can apply the natural logarithm to both sides of the inequality: $$-\frac{x^2 + y^2}{8} \geq -2.$$ Multiplying by -8, we obtain: $$x^2 + y^2 \leq 16$$. Hence, the region where the function is nonnegative is the disc centered at the origin (0,0) with radius 4.

Set up the integral for the volume calculation

To find the volume of the island, we should integrate the function $$z=e^{-\left(x^{2}+y^{2}\right) / 8}-e^{-2}$$ over the region determined in step 1. Since the region is a disc defined by the inequality $$x^2 + y^2 \leq 16$$, it's convenient to use polar coordinates to set up the integral. - First, convert the function to polar coordinate expression, where $$x = r \cos(\theta)$$ and $$y = r \sin(\theta)$$. Thus, $$z = e^{-\left(r^2\cos^2(\theta) + r^2\sin^2(\theta)\right) / 8}-e^{-2} = e^{-r^2 / 8}-e^{-2}$$. - Since we're integrating over a disc, the limits for $$r$$ will be from 0 to 4, and the limits for $$(\theta)$$ will be from 0 to $$2\pi$$. Now we can set up the integral for the volume as follows: $$V=\int_{0}^{2\pi} \int_{0}^{4} \left(e^{-r^2 / 8}-e^{-2}\right)r \, dr \, d\theta$$.

Evaluate the integral and find the volume

To find the volume, we need to evaluate the nested integrals. We will first integrate the function with respect to $$r$$. The antiderivative with respect to $$r$$ is: $$F(r) = -4e^{-r^2 / 8} - re^{-2}$$ Evaluate it from the limits 0 to 4, we get: $$F(4) - F(0) = -4e^{-\frac{16}{8}} - 4e^{-2} - (-4e^{-\frac{0}{8}} - 0) = -4(e^{-2} - e^0)$$. Now we need to integrate $$-4(e^{-2} - e^0)$$ with respect to $$\theta$$ over the interval from 0 to $$2\pi$$: $$V = \int_{0}^{2\pi} -4(e^{-2} -1)d\theta$$. Since the integrand does not depend on $$\theta$$, the integral is straightforward: $$V = -4(e^{-2} - 1)(2\pi - 0)$$. Finally, we get the volume of the island: $$V = -8\pi(e^{-2} - 1) \approx 15.3623$$.

Key concepts

Polar Coordinates

Polar coordinates offer a convenient way to describe points in a plane using a distance and an angle. Instead of using traditional Cartesian coordinates \((x, y)\), polar coordinates are expressed as \((r, \theta)\).

• **\(r\)** - Represents the distance from the origin to the point.
• **\(\theta\)** - Represents the angle measured from the positive x-axis.

For the problem of finding the volume of revolution, polar coordinates are especially helpful when dealing with circular regions like discs. In this scenario,

• The equations \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\) allow conversion from Cartesian to polar coordinates.
• In integration, the differential area element is represented as \(r \, dr \, d\theta\) which simplifies calculations in circular domains.

This way, working with functions involving squares of \(x\) and \(y\), particularly in discs, becomes more straightforward.

Double Integral

The double integral is a powerful tool for calculating volumes under surfaces. It involves two integrations, first with respect to one variable and then the other.

• A double integral over a region \(R\) can be expressed as \(\int \int_R f(x, y) \; dx \; dy\).

By converting regions to polar coordinates, the expression often becomes simpler, especially helpful for circular regions. For a region that's a disc centered at the origin, using polar coordinates:

• The double integral becomes \(\int_{0}^{2\pi} \int_{0}^{4} f(r, \theta) \; r \, dr \, d\theta\), where \(r\) and \(\theta\) replace \(x, y\) coordinates.

Evaluating the double integral involves:

• Choosing proper bounds for \(r\) and \(\theta\).
• Integrating first with respect to \(r\), then over \(\theta\).

Understanding these steps ensures a correct evaluation of the volume under complex surfaces.

Exponential Function

Exponential functions describe rapid growth or decay and are denoted as \(e^x\) where \(e\) is approximately 2.71828. They are prevalent in mathematical models, including geometry.

• In the given island surface, the function \(z = e^{-\frac{x^2 + y^2}{8}} - e^{-2}\) includes an exponential decay term.

This describes how the function decreases as \((x^2 + y^2)\) increases.

• Here, \(e^{-\frac{x^2 + y^2}{8}}\) implies a significant reduction in value as one moves away from the origin, modeling a hill-like surface with a peak at the center.
• The subtracted constant \(e^{-2}\) determines the lowest value \(z\) can attain, ensuring the function remains non-negative over a defined region.

Grasping the behavior of exponential functions is crucial to understanding how surfaces change over the x-y plane.

Disc Region

The disc region is critical to this problem as it defines the boundaries of the island's surface. In this scenario, the region is determined by \(x^2 + y^2 \leq 16\).

• This inequality represents a circle with radius 4 centered at the origin.

When analyzing and integrating over circular areas, using a disc region simplifies calculations tremendously by leveraging symmetry.

• The region in polar coordinates is \(0 \leq r \leq 4\) and \(0 \leq \theta \leq 2\pi\).

This circular description allows for easier setting of limits in integrals, especially when finding volumes under curves.

• Using polar coordinates to integrate within a disc ensures accuracy and simplicity in geometric problems.

Disc regions often appear in problems involving rotational symmetry, making them a fundamental aspect of analyzing such surfaces.