Question

Find the volume of the following solid regions. The solid above the region \(R=\\{(x, y): 0 \leq x \leq 1\) \(0 \leq y \leq 1-x\\}\) bounded by the paraboloids \(z=x^{2}+y^{2}\) and \(z=2-x^{2}-y^{2}\) and the coordinate planes in the first octant

Short answer

Question: Find the volume of the solid region bounded by the paraboloids \(z_1 = x^2 + y^2\) and \(z_2 = 2 - x^2 - y^2\), and the limits of the region R, given by \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\). Answer: The volume of the solid region is \(\frac{1}{18}\).

Step-by-step solution

Determine the Limits of Integration

Using the region R, we can see that the limits of integration are \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\).

Set Up the Double Integral

To find the volume of the solid bounded by the surfaces given, we need to subtract the z-values of the paraboloids and set up a double integral over the region R. We will integrate \(((2 - x^2 - y^2) - (x^2 + y^2))\) with respect to x and y using the integration limits from Step 1. The double integral is: $$\int_{0}^{1} \int_{0}^{1-x} (2 - 2x^2 - 2y^2) dy dx$$

Integrate with respect to y

Now we need to integrate the inner integral with respect to y: $$\int_{0}^{1} \Big[ y - 2x^{2}y - \frac{2}{3}y^{3} \Big]_{0}^{1-x} dx$$ Substitute the upper and lower limits for y: $$\int_{0}^{1} \Big[ (1-x) - 2x^2(1-x) - \frac{2}{3}(1-x)^3 \Big] dx$$

Integrate with respect to x

Now we need to integrate the outer integral with respect to x: $$\int_{0}^{1} (1 - x - 2x^2 + 2x^3 - \frac{2}{3} + \frac{4}{3}x - \frac{2}{3}x^2 + \frac{2}{3}x^3) dx$$ We can simplify the integrand: $$\int_{0}^{1} (1 - x - \frac{8}{3}x^2 + \frac{8}{3}x^3) dx$$ Now, we can integrate this expression with respect to x: $$\Big[x - \frac{1}{2}x^2 - \frac{8}{9}x^3 + \frac{2}{3}x^4 \Big]_{0}^{1}$$

Evaluate the Integral and Find the Volume

Finally, substitute the values of the limits of integration x = 1 and x = 0: $$\Big[ 1 - \frac{1}{2} - \frac{8}{9} + \frac{2}{3} \Big] - \Big[0 - 0 - 0 + 0 \Big]$$ Simplify the expression to find the volume: $$= -\frac{1}{2} + 1 - \frac{8}{9} + \frac{2}{3} = \frac{1}{18}$$ The volume of the solid region is \(\frac{1}{18}\).

Key concepts

Double Integrals

Double integrals allow us to calculate the volume of a solid region in space. They extend the concept of single integrals to two dimensions, making it possible to evaluate how a function behaves over a region in the xy-plane. In our problem, we are calculating the volume between two paraboloids, which involves a double integral over the region defined by the limits for x and y.
To set up a double integral, we specify limits for both variables and then integrate the function representing the height between the surfaces. Here, the function we integrate is \(2 - 2x^2 - 2y^2\), and we integrate it twice: once with respect to y and then with respect to x.

• The inner integral accounts for variations along the y-axis.
• The outer integral sums these results along the x-axis.

Limits of Integration

Defining the limits of integration is a crucial step in evaluating double integrals. These limits tell us the boundaries over which we integrate. Let’s break down how the limits are chosen for our problem:
First, we consider the region R, bounded by constraints \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\).

• The limit for x is straightforward: from 0 to 1.
• For each fixed x, y ranges from 0 to the line \(y = 1-x\), which forms a right triangle in the xy-plane.

These limits tell us that our integration region is the triangular area under the line, within the specified boundaries.
Choosing the correct limits ensures accuracy in calculation and proper coverage of the region.

Paraboloids

Paraboloids are surfaces that have parabolic cross-sections. In this problem, they’re represented by the equations \(z = x^2 + y^2\) and \(z = 2 - x^2 - y^2\). These equations describe two upward and downward opening shapes respectively.
Understanding the geometry of these paraboloids helps us set up our integrals:

• The first paraboloid, \(z = x^2 + y^2\), opens upwards from the xy-plane.
• The second paraboloid, \(z = 2 - x^2 - y^2\), opens downwards and is above the first for the region in question.

Visualizing these shapes as layers helps us realize what the double integral calculates: the volume of space between these two surfaces over the given region.

Coordinate Planes

The coordinate planes, such as the xy-plane, help define boundaries and provide fixed reference points for evaluating integrals. In our specific problem, the solid lies in the first octant, meaning:

• All coordinates are non-negative: \(x, y, z \geq 0\).
• The xy-plane acts as both a boundary and a reference for locating the intersecting paraboloids.

This restriction helps set up the problem context, simplifying the range of possible x and y values to consider. This focus on the first octant is crucial because it defines the context where the surface is being integrated. Understanding how the coordinate planes interact with the regions helps visualize the problem and guide integral setup.