Question

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$ z=x y, z=0, y=0, y=4, x=0, x=1 $$

Short answer

The volume of the solid bounded by the given equations is 1 cubic units.

Step-by-step solution

Determine the Range for Integration

If you look at the equation constraints provided, the region over which we will be integrating z=xy is the rectangle defined by y from 0 to 4 and x from 0 to 1. This will serve as the range for our double integral.

Set up Double Integral

We can set up the double integral for this scenario as follows: \[\int_{0}^{1}\int_{0}^{4} xy \, dy\, dx\]. Recall that in this double integral representation, dy indicates that y is to be integrated first, then followed by integration with respect to x.

Evaluate the Inner Integral

The inner integral refers to the integration with respect to y. We find the antiderivative of xy with respect to y, treating x as a constant, which is \(\frac{1}{2}xy^2\). We evaluate this from y=0 to y=4 which gives \(2x\)-0 = 2x.

Evaluate the Outer Integral

The outer integral now refers to the integration with respect to x. We integrate 2x from x=0 to x=1 to give \(x^2\) evaluated at 1 and 0, which gives 1-0 = 1.

Key concepts

Solid Volume Using Double Integrals

The calculation of the volume of a solid using double integrals is a powerful tool in multivariable calculus. It allows us to find the volume of a three-dimensional region by integrating a function over a two-dimensional region. The volume of a solid can be represented as:
\[\begin{equation} \text{Volume} = \int\int \text{Density Function} \,dA \end{equation}\]
Here, the density function is usually the function that defines the height of the solid above the xy-plane. In our example, the density function is given by the equation \( z = xy \). This represents a surface and the volume we are after is under this surface and above the xy-plane. The volume is therefore calculated by integrating the function \( xy \) over the rectangular region defined by \( 0 \leq y \leq 4 \) and \( 0 \leq x \leq 1 \).

The region of integration is crucial, as it defines the 'base' of the solid whose volume we are calculating. For different shapes of bases or different equations, it's important to correctly identify and set the limits of integration to ensure the entire base area is covered and no extraneous parts are included.

• Base Region: The part of the xy-plane over which integration is performed.
• Density Function: The function representing height, \( z \) in our case.
• Volume: The three-dimensional measure we're seeking to compute.

Setting Up Double Integrals

Correctly setting up a double integral is an essential step in performing solid volume calculations. To establish the limits of integration, we must first understand the region of integration. For solids bounded by simple geometric regions, defining these limits is relatively straightforward.

In our textbook problem, the integral is set up over a rectangle in the xy-plane, with the bounds for y ranging from 0 to 4 and the bounds for x ranging from 0 to 1. The integral is organized as:\[\begin{equation} \int_{0}^{1}\int_{0}^{4} xy \, dy\, dx \end{equation}\]
This double integral can be visualized as accumulating infinitesimal rectangles with height \( xy \) and base area \( dA = dx \cdot dy \). To accurately set up a double integral, one must:

• Determine the region of integration in the xy-plane.
• Choose the order of integration (which variable to integrate first).
• Write down the double integral with the appropriate limits matching the region and order of integration.

For more complex shapes or boundaries, one might need to apply more advanced techniques to find the appropriate limits, which may involve splitting the region into simpler sub-regions or using different coordinate systems such as polar coordinates.

Evaluating Inner and Outer Integrals

Once we have established our double integral, we progress to evaluating it, starting with the inner integral and then proceeding to the outer integral. The inner integral involves fixing one variable and integrating with respect to the other. In the example problem, we treat \( x \) as a constant and integrate with respect to \( y \), finding the antiderivative of the function \( xy \) which is:

\[\begin{equation} \int xy \, dy = \frac{1}{2}xy^2 \end{equation}\]
After evaluating this from \( y = 0 \) to \( y = 4 \) for each \( x \) and simplifying, we are left with the function \( 2x \) which is then used in the outer integral.

The outer integral is now the simpler task of integrating the result of the inner integral with respect to the outer variable. In our case, this is \( x \) and we integrate from 0 to 1:\[\begin{equation} \int 2x \,dx = x^2 \end{equation}\]
Finally, we evaluate this from \( x = 0 \) to \( x = 1 \) and find the volume of the solid to be 1 cubic unit.

The process of evaluating double integrals often involves a systematic approach:

• Solve the inner integral first, treating the outer variable as constant.
• Evaluate the result of the inner integral at the given bounds.
• Proceed with the outer integral using the evaluated result from the inner integral.
• Evaluate the final result at the bounds of the outer variable.

Doing each step carefully ensures an accurate calculation of the volume of the solid. It's important to understand how the process relates to summing up the volumes of infinitesimal slices of the solid to achieve the final aggregate volume.