Question

Find the volume in the first octant bounded by the paraboloid \(z=1-x^{2}-y^{2},\) the plane \(x+y=1,\) and all three coordinate planes.

Short answer

The volume is \( \frac{1}{6} \).

Step-by-step solution

Identify the region of integration

First, identify the region in the first octant bounded by the paraboloid, the plane, and the coordinate planes. Since we are in the first octant, this means that all variables are non-negative. The paraboloid is given by \(z=1-x^{2}-y^{2}\), and the plane by \(x+y=1\).

Set up the limits of integration

We need to find the intersection of the paraboloid and the plane to determine the limits of integration. Plugging the plane equation \(x+y=1\) into the paraboloid equation: \[z = 1 - x^2 - y^2\ \rightarrow z = 1 - x^2 - (1-x)^2 = 2x - 1\]The limits for \(x\) and \(y\) will range from 0 to 1, as determined by the plane equation and the boundaries in the first octant.

Express the volume integral

Express the volume integral in terms of \(x\), \(y\) and \(z\). The volume \(V\) is given by:\[V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x^2-y^2} dz\, dy\, dx\]

Compute the innermost integral

Evaluate the innermost integral with respect to \(z\):\[\int_{0}^{1-x^2-y^2} dz = [z]_{0}^{1-x^2-y^2} = 1 - x^2 - y^2\]

Compute the second integral

Substitute the result of the innermost integral, and then evaluate the integral with respect to \(y\):\[\int_{0}^{1-x} (1 - x^2 - y^2) dy\]This integral can be separated:\[(1 - x^2) \int_{0}^{1-x} dy - \int_{0}^{1-x} y^2 dy\]\[(1 - x^2)(1-x) - \int_{0}^{1-x} y^2 dy\]

Compute the \(y^2\) term integral

Now evaluate the integral of \(y^2\):\[\int_{0}^{1-x} y^2 dy = [\frac{y^3}{3}]_{0}^{1-x} = \frac{(1-x)^3}{3}\]

Simplify and combine terms

Combine the results of the integrals and simplify:\[(1 - x^2)(1 - x) - \frac{(1-x)^3}{3} = 1-x-x^2+x^3 - \frac{(1-x)^3}{3}\]Further expanding, we get:\[1 - x - x^2 + x^3 - \frac{1 - 3x + 3x^2 - x^3}{3} = 1 - x - x^2 + x^3 - \frac{1}{3} + x - x^2 + \frac{x^3}{3}\]

Final integration with respect to \(x\)

Finally, integrate with respect to \(x\):\[\int_{0}^{1} (1 - x - x^2 + x^3 - \frac{1}{3} - x + \frac{x^2}{3} + \frac{x^3}{3}) dx\]Upon simplification: \[\int_{0}^{1} (\frac{2}{3} - x - \frac{2}{3}x^2 + \frac{2}{3}x^3) dx\]Evaluate each term separately and sum the results:\[\frac{2}{3}x - \frac{x^2}{2} - \frac{2}{3}\frac{x^3}{3} + \frac{2}{3}\frac{x^4}{4}\]Evaluate these from 0 to 1.

Simplify and find the volume

Evaluating our integral gives:\[(\frac{2}{3} - \frac{1}{2} - \frac{2}{9} + \frac{2}{12})\]This simplifies to approximately \[\frac{1}{6}\]Hence, the volume \(V\) is \(\frac{1}{6}\).

Key concepts

Paraboloid

A paraboloid is a 3-dimensional surface that can be visualized by rotating a parabola around its axis. In this exercise, the given paraboloid is described by the equation \(z=1-x^2-y^2\). This means that every point on this surface satisfies this relationship between \(x\), \(y\), and \(z\). The paraboloid opens downwards because the coefficients of \(x^2\) and \(y^2\) are negative.

The surface intersects the plane \(z=1\) at the origin \((x=0, y=0)\) and extends downward, forming a bowl shape. Understanding the shape of a paraboloid can greatly help in visualizing the region bounded by it, assisting in the process of setting up the integration limits.

Double Integrals

Double integrals allow us to compute volumes under surfaces or areas enclosed by curves in a two-dimensional plane. In our exercise, we use a double integral to find the volume underneath the paraboloid and above the region bounded by the plane \(x+y=1\).

When setting up a double integral, it's essential to identify the region of integration properly. Here, it involves integrating over a region in the \(xy\)-plane, and then integrating with respect to \(z\). We first compute an inner integral with respect to one variable, then an outer integral with respect to the other variable.

The integral that represents the volume is given by:
\[V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x^2-y^2} dz \, dy \, dx\]
The limits and order of integration find crucial importance to accurately capture the enclosed volume.

Integration Limits

Setting accurate integration limits is paramount in computing multiple integrals. In this problem, we are restricted to the first octant where all coordinates are non-negative (\(x \geq 0\), \(y \geq 0\), \(z \geq 0\)).

For the inner integral with respect to \(z\), the upper limit is determined by the paraboloid equation: \(z = 1 - x^2 - y^2\). For the middle integral with respect to \(y\), the limits are from \(0\) to \(1 - x\), where \(1-x\) comes from the plane equation \(x + y = 1\). The outer integral with respect to \(x\) ranges from \(0\) to \(1\) since \(x\) and \(y\) must be non-negative and cannot exceed the value where they sum to one.

Visualizing the region bounded by the paraboloid and the coordinate planes helps to correctly determine these limits.

Coordinate Planes

Coordinate planes are the planes formed by setting one of the coordinates \((x, y, \text{or} \text{z})\) to zero. In this exercise, the volume is bounded by the three coordinate planes \(x=0\), \(y=0\), and \(z=0\).

These planes are essential as they define the boundaries of our region within the first octant. The plane \(x=0\) is the \(yz\)-plane, \(y=0\) is the \(xz\)-plane, and \(z=0\) is the \(xy\)-plane. These coordinate planes ensure that we only consider the part of the paraboloid and plane that lies in the first octant.

Therefore, setting up your integration within these boundaries guarantees you compute the volume of the correct region.