Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use triple iterated integrals to find the indicated quantities. Volume of the solid in the first octant bounded by \(y=2 x^{2}\) and \(y+4 z=8\)

Step-by-Step Solution

Answer

The volume is computed by setting up and evaluating the triple integral, \( V = \int_{0}^{2} \int_{2x^2}^{8} \int_{0}^{\frac{8-y}{4}} dz\, dy\, dx \).

Step 1: Set Up the Problem

The problem requires us to find the volume of a solid contained in the first octant. The solid is bounded by the surface given by the equation \(y = 2x^2\) and the plane described by \(y + 4z = 8\). The first octant implies that \(x, y, z \geq 0\).

Step 2: Determine the Limits of Integration for z

First, solve the equation of the plane for \(z\): \[ z = \frac{8 - y}{4} \]Since we're only considering the region in the first octant (where \(z \geq 0\)), \(z\) is bounded between 0 and \(\frac{8-y}{4}\).

Step 3: Determine the Limits of Integration for y

Next, consider \(y\) in terms of \(x\) by using the equation \(y = 2x^2\). This gives us the lower bound of \(y\) as \(0\) and upper bound as \(8\), from solving \(y = (8-y)/4\), we find that \(y = 8/5\). Thus, the limits of \(y\) are from \(2x^2\) to 8.

Step 4: Determine the Limits of Integration for x

The limits for \(x\) are determined by the range in which \(y\) is non-negative. Since \(y = 2x^2\), the range for \(x\) is between \(0\) and \(\sqrt{\frac{8}{2}} = 2\) such that \(y \leq 8\).

Step 5: Set Up the Integral

Using the determined limits, set up the triple iterated integral to find the volume:\[V = \int_{0}^{2} \int_{2x^2}^{8} \int_{0}^{\frac{8-y}{4}} dz\, dy\, dx\]

Step 6: Evaluate the Integral Layer by Layer

First, integrate with respect to \(z\):\[V = \int_{0}^{2} \int_{2x^2}^{8} \left[ \frac{8-y}{4} \right]_{0}^{\frac{8-y}{4}} dy\, dx\]This simplifies to:\[\int_{0}^{2} \int_{2x^2}^{8} \left( \frac{8-y}{4} \right) dy\, dx\]

Step 7: Integrate with Respect to y

Now, integrate with respect to \(y\):\[\int_{0}^{2} \left[ \frac{8-y}{4} \cdot y - \frac{1}{8}y^2 \right]_{2x^2}^{8} dx\]When evaluated, this step yields a function solely in \(x\).

Step 8: Integrate with Respect to x

Finally, integrate the previous result with respect to \(x\):Evaluate \[\int_0^2 \left( \text{evaluated function in terms of } x \right) dx\]This will give you the volume of the solid.

Step 9: Compute and Simplify

Perform the arithmetic to find a numerical value, complete all substitutions, simplifications, and evaluate the integrals, either by hand or using a computational system as needed.

Key Concepts

volume of solid

In multivariable calculus, finding the volume of a solid bounded by a set of surfaces requires setting up and evaluating iterated integrals. The volume is essentially the three-dimensional space occupied by the solid. To compute it using triple integrals, we apply integration over a suitable domain defined by the bounding surfaces. Each integral represents an accumulation over one of the dimensions.
By iterating integrals, we begin from the innermost integral, which corresponds to the dimension resolved last, and proceed outward. Evaluating a triple integral provides us with the total volume contained within the specified boundaries.

first octant

The first octant in three-dimensional space refers to the portion where all coordinates are non-negative, which are typically denoted as \(x, y, z \geq 0\). This means that each of the coordinates is greater than or equal to zero.
When determining the volume of a solid in this specific region of space, the boundaries of integration must respect these non-negativity constraints. This is crucial because any solution derived for a problem in the first octant is physically meaningful only if it stays within this region. Therefore, it affects how you define limits for each variable in your integral setup.

bounds of integration

To find the bounds of integration for a triple iterated integral, we determine how the region of integration is constrained by the problem's surfaces. These bounds specify where exactly in space you're interested in calculating volume or other quantities.
In the context of finding volume, the bounds for each variable are established by evaluating the limits in terms of other variables progressively. For instance, solve the equations defining bounding surfaces (like planes or curves) for each variable sequentially: in our problem, start with the most interior variable, moving outward.

• For \(z\), the bounds were found from the plane \(y + 4z = 8\).
• For \(y\), limits were identified from the surface \(y = 2x^2\).
• Finally, \(x\)'s range is determined through the feasible \(y\) values attainable within the first octant.

Proper bounds ensure the integral samples the entire solid within the defined limits.

multivariable calculus

Multivariable calculus extends the principles of single-variable calculus to more than one variable. This branch explores derivatives, integrals, and other concepts applied to functions of multiple variables. This is vital for understanding how changes in different variables collectively influence a function's behavior.
One key application is the computation of triple integrals to determine volumes or masses in higher dimensions. These integrals operate over regions defined by constraints on multiple variables. In our original problem, the triple integral reflects interactions among variables \(x, y,\) and \(z\) to find volume efficiently.

• Partial differentiation involves computing derivatives with respect to one variable at a time.
• Triple integrals require integrating a function over a three-dimensional region, accounting for dependencies between variables.

By mastering these tools, we can solve complex real-world problems involving spatial dimensions.