Question

Sketch the indicated solid. Then find its volume by an iterated integration. Solid in the first octant bounded by the cylinder \(y=x^{2}\) and the planes \(x=0, z=0\), and \(y+z=1\)

Short answer

The volume of the solid is \(\frac{1}{5}\).

Step-by-step solution

Understand the Region in the First Octant

The solid is in the first octant, which means that the coordinates \((x, y, z)\) are non-negative. The boundary conditions given are: the cylinder \(y = x^2\), the plane \(y + z = 1\), and the planes \(x = 0\) and \(z = 0\). We need to focus on these conditions to determine the limits of integration.

Sketch the Region

Visualize the region of integration. Start by plotting the cylinder \(y = x^2\), which is a parabola in the \(xy\)-plane. The plane \(y + z = 1\) intersects the cylinder. Additionally, the region is bounded by \(x = 0\) (the \(yz\)-plane) and \(z = 0\) (the \(xy\)-plane). The sketch will resemble an area under the curve \(y = x^2\) and below the line \(z = 1-y\) within the first octant.

Determine Limits for Integration

For the integration setup, consider how each variable gets bounded:- The limits for \(x\) range from \(0\) to \(1\) since \(y = x^2\) and lies in the first octant where \(x\) must be non-negative and cannot exceed \(1\) (as given by \(y \leq 1\)).- For each \(x\), \(y\) ranges from \(x^2\) to \(1\).- For given \(x\) and \(y\), \(z\) ranges from 0 to \(1-y\).

Set up the Double Integral for Volume

With the region and limits known, the volume \(V\) of the solid can be found by using an iterated integral:\[V = \int_{0}^{1} \int_{x^2}^{1} \int_{0}^{1-y} dz \ dy \ dx\]

Evaluate the Inner Integral with Respect to z

Integrate with respect to \(z\):\[\int_{0}^{1-y} dz = z \Big|_{0}^{1-y} = 1-y\]Thus, the integral becomes:\[\int_{0}^{1} \int_{x^2}^{1} (1-y) \ dy \ dx\]

Evaluate the Middle Integral with Respect to y

Now integrate \(1-y\) with respect to \(y\):\[\int_{x^2}^{1} (1-y) dy = \left[ y - \frac{y^2}{2} \right]_{x^2}^{1} = \left(1 - \frac{1}{2}\right) - \left(x^2 - \frac{x^4}{2}\right)\]Simplifying, we have:\[\frac{1}{2} - x^2 + \frac{x^4}{2}\]Thus, the integral becomes:\[\int_{0}^{1} \left(\frac{1}{2} - x^2 + \frac{x^4}{2}\right) dx\]

Evaluate the Outer Integral with Respect to x

Integrate with respect to \(x\):\[\int_{0}^{1} \left(\frac{1}{2} - x^2 + \frac{x^4}{2}\right) dx = \frac{x}{2} - \frac{x^3}{3} + \frac{x^5}{10} \Big|_{0}^{1} = \frac{1}{2} - \frac{1}{3} + \frac{1}{10}\]Simplifying:\[\frac{5}{10} - \frac{10}{30} + \frac{3}{30} = \frac{1}{5}\]Hence, the volume \(V = \frac{1}{5}\).

Final Step: Conclude with the Volume of Solid

After evaluating all integrals stepwise, the volume of the solid bounded by the cylinder and planes in the first octant is \(\frac{1}{5}\).

Key concepts

Iterated Integration

Iterated integration is a powerful method used to calculate the volume of a three-dimensional shape or solid. It involves breaking down a multi-variable integral into a series of single-variable integrals. In this exercise, the goal is to find the volume of a solid within certain boundaries using iterated integration.

The process begins by considering the order in which integration should be performed. Typically, one integrates with respect to one variable, while holding the others constant, then moves on to the next variable. This exercise requires integrating with respect to the variables: \(z\), \(y\), and \(x\) sequentially. By doing so, the task of evaluating a complex triple integral is transformed into a series of easier single integrals, making the problem more manageable.

Cylinder in First Octant

In this problem, the solid is situated in the first octant of a Cartesian coordinate system. The first octant is the region where all coordinate values \((x, y, z)\) are non-negative. This particular solid is bounded by a cylindrical surface defined by \(y = x^2\).

This cylinder is a parabolic surface extending infinitely along the \(z\)-axis. However, for our problem, it is also confined by other planes, creating a finite region in the first octant.

• The intersection of the cylinder with the plain \(y + z = 1\) provides the uppermost bound of the solid.
• The planes \(x = 0\) and \(z = 0\) serve as additional constraints, indicating the solid does not extend into negative directions along the \(x\) and \(z\) axes.

Understanding these spatial boundaries is crucial for accurately determining the limits for integration.

Limits of Integration

Defining the correct limits of integration is critical for solving the volume of the solid accurately. These limits are determined based on the nature of the bounding surfaces:

• The variable \(x\) ranges from \(0\) to \(1\) because \(x\) is non-negative and cannot exceed the intersection of \(y = x^2\) with \(y = 1\).
• For each specific value of \(x\), \(y\) varies between \(x^2\) and \(1\). This range is due to the fact that \(y = x^2\) is the base, and \(y + z = 1\) limits \(y\) to a maximum of \(1\).
• Finally, \(z\) extends from \(0\) to \(1-y\) for each pair of \(x, y\) values, acknowledging that \(y + z = 1\) serves as an upper constraint.

These boundaries ensure that each point inside the solid is encapsulated by the triple integral, maximizing accuracy in the calculation of volume.

Triple Integral

A triple integral extends the concept of integration to three dimensions and is used for calculating volumes when geometric constraints are known.

In this problem, the volume \(V\) of the solid is computed using a triple integral of the form:\[V = \int_{0}^{1} \int_{x^2}^{1} \int_{0}^{1-y} dz \ dy \ dx\]

Here's how it works:

• The innermost integral \(\int_{0}^{1-y} dz\) is evaluated first, summing along the \(z\)-axis direction while \(x\) and \(y\) remain constants.
• Next, the middle integral \(\int_{x^2}^{1} \) is performed with respect to \(y\), considering the already determined constraint \(1-y\).
• The outermost integral \(\int_{0}^{1} \) ranges across \(x\), aggregating all previous results.

Through this layered integration process, the true volume of the solid, considering all boundary restrictions, is calculated as \(\frac{1}{5}\). This application of a triple integral is a sophisticated way to handle complex geometric shapes.