Chapter 14: Problem 30
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. $$z=100-4\left(x^{2}+y^{2}\right)$$
Short Answer
Expert verified
Answer: The volume of the island is $$V = 625(2\pi)$$ cubic units.
Step by step solution
01
Determine the region of integration
First, we need to find the region over which the function is nonnegative. We have $$z = 100 - 4(x^2 + y^2) \geq 0$$ which implies that
$$x^2 + y^2 \leq \frac{100}{4} = 25$$
This inequality represents the region inside a circle of radius 5 centered at the origin in the xy-plane.
02
Set up the double integral
Now we can set up the double integral over the region to find the volume. Using polar coordinates, we have $$x = r\cos(\theta)$$ and $$y = r\sin(\theta)$$, so the region in polar coordinates is defined by \(0 \leq r \leq 5\) and \(0 \leq \theta \leq 2\pi\). The double integral can then be expressed as
$$
V = \int_{0}^{2\pi} \int_{0}^{5} (100 - 4(r^2\cos^2(\theta) + r^2\sin^2(\theta))) r dr d\theta
$$
Since \(cos^2(\theta) + sin^2(\theta) = 1\), the integral becomes
$$
V = \int_{0}^{2\pi} \int_{0}^{5} (100 - 4r^2) r dr d\theta
$$
03
Evaluate the integral
Now we can evaluate the integral as follows:
$$
V = \int_{0}^{2\pi} \left[ \int_{0}^{5} (100r - 4r^3) dr \right] d\theta
$$
First, we integrate with respect to r:
$$
\int_{0}^{5} (100r - 4r^3) dr = \left[\frac{100}{2}r^2 - \frac{4}{4}r^4 \right]_0^5 \\
= \left[\frac{100}{2}(5^2) - \frac{4}{4}(5^4) \right] \\
= 1250 - 625 = 625
$$
Now, integrating with respect to \(\theta\):
$$
V = \int_{0}^{2\pi} 625 d\theta = 625\left[\theta\right]_0^{2\pi} = 625(2\pi)
$$
So the volume of the island is $$V = 625(2\pi)$$ cubic units.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integrals
Double integrals are a way to calculate the volume under a surface in a two-dimensional region. They extend the idea of a single integral to functions of two variables. When performing a double integral, you integrate one variable at a time. This often involves setting up a function over a region in the plane and integrating over that region.
In our exercise, we used a double integral to find the volume of the island defined by the surface equation. We first determined the region over which we need to integrate, and then set up the integral using the function given.
The process involves:
In our exercise, we used a double integral to find the volume of the island defined by the surface equation. We first determined the region over which we need to integrate, and then set up the integral using the function given.
The process involves:
- Setting up the bounds of the region of integration.
- Aligning each variable's bounds with the region described in the xy-plane.
- Performing integration with respect to one variable and then the other.
Polar Coordinates
Polar coordinates are a system for representing points in a plane using the distance from a reference point and an angle from a reference direction. This system is particularly useful for circular or radial symmetry problems. In polar coordinates, a point is represented as \(r, \theta\) where \(r\) is the radius and \(\theta\) is the angle.
For many problems in calculus that involve circles or circular regions, converting from Cartesian coordinates \(x, y\) to polar coordinates simplifies the computation. This is because it leverages the symmetry of the problem, turning complex problems into simpler ones.
Here's how polar coordinates were used in the exercise:
For many problems in calculus that involve circles or circular regions, converting from Cartesian coordinates \(x, y\) to polar coordinates simplifies the computation. This is because it leverages the symmetry of the problem, turning complex problems into simpler ones.
Here's how polar coordinates were used in the exercise:
- Converting \(x\) and \(y\) to \(r\cos(\theta)\) and \(r\sin(\theta)\).
- Updating the region's bounds to \(0 \leq r \leq 5\) and \(0 \leq \theta \leq 2\pi\).
- Using the transformation \(dA = r\,dr\,d\theta\) in the integral setup.
Region of Integration
Determining the region of integration is a foundational step in setting up double integrals. The region where the function is nonnegative determines where we integrate.
In our exercise, the given function was \(z = 100 - 4(x^2 + y^2)\), and we needed to identify where this function is greater than or equal to zero. To do this, we solved the inequality \(x^2 + y^2 \leq 25\), which describes a region that is a circle with a radius of 5, centered at the origin.
Here are the steps involved:
In our exercise, the given function was \(z = 100 - 4(x^2 + y^2)\), and we needed to identify where this function is greater than or equal to zero. To do this, we solved the inequality \(x^2 + y^2 \leq 25\), which describes a region that is a circle with a radius of 5, centered at the origin.
Here are the steps involved:
- Solve for the values of \(x\) and \(y\) that satisfy the inequality.
- Draw or visualize the corresponding region (here a circle).
- Ensure the region is appropriately bounded for integration.
Circle in xy-plane
A circle in the xy-plane is defined as the set of all points that are a fixed distance from a central point, typically the origin. Mathematically, it's described by \(x^2 + y^2 = r^2\), where \(r\) is the radius.
In the context of the exercise, knowing the circle helps define the region of integration. The function above formed a circular domain due to the quadratic terms in \(x\) and \(y\) when solving the inequality \(x^2 + y^2 \leq 25\). This led us to a circle of radius 5 centered at the origin.
Key insights on circles for integration problems:
In the context of the exercise, knowing the circle helps define the region of integration. The function above formed a circular domain due to the quadratic terms in \(x\) and \(y\) when solving the inequality \(x^2 + y^2 \leq 25\). This led us to a circle of radius 5 centered at the origin.
Key insights on circles for integration problems:
- Recognize the equation of a circle and relate it to the region of integration.
- Use circular symmetry to choose the right coordinate system for integration (like polar coordinates).
- It helps in visualizing the problem and understanding the relationship between geometry and calculus.