Find The Volume Of The Solid Bounded By The Xy- Plane And The Surfaces X + Y=144 And Z=x+y. ______

Find The Volume Of The Solid Bounded By The Xy- Plane And The Surfaces X + Y=144 And Z=x+y.

Understanding how to find the volume of a solid bounded by specific surfaces and planes is a fundamental problem in multivariable calculus. This type of problem involves setting up and evaluating a double integral over a region in the xy-plane, where the limits are defined by the given surfaces. In this article, we will explore the step-by-step process to determine the volume of the solid bounded by the xy-plane, the plane x + y = 144, and the surface z = x + y. We will cover the geometric interpretation, the setup of the integral, methods for evaluating it, and practical tips to ensure accurate calculations. This comprehensive guide aims to clarify the concepts and provide a thorough understanding suitable for students, educators, and anyone interested in calculus applications.

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Understanding the Geometric Configuration

Before delving into the calculations, it’s essential to visualize the problem and understand the geometric shapes involved.

Surfaces and Planes Involved

• XY-plane (z=0): The base of the solid, lying flat in the xy-plane.
• Plane x + y = 144: A straight plane inclined in 3D space, intercepting axes at (144, 0, 0) and (0, 144, 0).
• Surface z = x + y: A plane that rises above the xy-plane, with z increasing linearly as x and y increase.

Region of Interest in the XY-Plane

The projection of the solid onto the xy-plane is a region bounded by:

• The coordinate axes (x=0, y=0).
• The line x + y = 144, which forms a straight boundary.

The region in the xy-plane is therefore a right triangle with vertices at:

• (0, 0)
• (144, 0)
• (0, 144)

This triangular region will be the domain of integration.

Visualizing the Solid

• The bottom surface is the xy-plane (z=0).
• The top surface is the plane z = x + y.
• The vertical boundaries are the coordinate axes and the line x + y = 144.

The solid resembles a three-dimensional wedge rising from the xy-plane up to the plane z = x + y, bounded horizontally by the triangle in the xy-plane.

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Setting Up the Volume Integral

The key to solving the problem is to express the volume as a double integral over the region R in the xy-plane:

\[
V = \iint_R \text{(height)} \, dA
\]

where the height at each point (x, y) is given by the surface z = x + y, since the solid extends from the xy-plane (z=0) up to z = x + y.

Formulating the Integral

• The height of the solid at a point (x, y) is \( z = x + y \).
• Therefore, the volume is:

\[ V = \iint_R (x + y) \, dA \]

• The region R is the triangle bounded by:
• x = 0
• y = 0
• x + y = 144

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Choosing the Order of Integration

Two common options are:

• Integrate with respect to y first, then x.
• Integrate with respect to x first, then y.

Given the boundary x + y = 144, it is convenient to integrate with respect to y first:

• For a fixed x, y varies from 0 to 144 - x.
• x varies from 0 to 144.

Thus, the limits are:

\[
0 \leq x \leq 144
\]
\[
0 \leq y \leq 144 - x
\]

The double integral becomes:

\[
V = \int{x=0}^{144} \int{y=0}^{144 - x} (x + y) \, dy \, dx
\]

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Calculating the Double Integral

Now, we proceed to evaluate the integral step-by-step.

Inner Integral

\[
Ix = \int{0}^{144 - x} (x + y) \, dy
\]

Since x is treated as a constant during the inner integration, we integrate:

\[
Ix = \int{0}^{144 - x} x \, dy + \int_{0}^{144 - x} y \, dy
\]

Calculating each term separately:

• \(\int_{0}^{144 - x} x \, dy = x \cdot (144 - x)\)


• \(\int_{0}^{144 - x} y \, dy = \frac{1}{2} (144 - x)^2\)

Therefore,

\[
I_x = x (144 - x) + \frac{1}{2} (144 - x)^2
\]

Outer Integral

The volume is then:

\[
V = \int_{0}^{144} \left[ x (144 - x) + \frac{1}{2} (144 - x)^2 \right] dx
\]

Simplify the integrand:

\[
V = \int_{0}^{144} \left[ 144x - x^2 + \frac{1}{2} (144 - x)^2 \right] dx
\]

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Evaluating the Integral

Break the integral into manageable parts:

\[
V = \int{0}^{144} 144x \, dx - \int{0}^{144} x^2 \, dx + \frac{1}{2} \int_{0}^{144} (144 - x)^2 \, dx
\]

Calculate each term separately.

First Term: \(\int_{0}^{144} 144x \, dx\)

\[
= 144 \cdot \frac{x^2}{2} \bigg|_{0}^{144} = 144 \cdot \frac{(144)^2}{2}
\]

Calculate:

\[
(144)^2 = 20736
\]

So,

\[
144 \cdot \frac{20736}{2} = 144 \times 10368 = 1,492,992
\]

Second Term: \(\int_{0}^{144} x^2 \, dx\)

\[
= \frac{x^3}{3} \bigg|_{0}^{144} = \frac{(144)^3}{3}
\]

Calculate \(144^3\):

\[
144^3 = 144 \times 144 \times 144 = (144 \times 144) \times 144 = 20736 \times 144
\]

\[
20736 \times 144 = (20000 + 736) \times 144 = 20000 \times 144 + 736 \times 144 = 2,880,000 + 106,304 = 2,986,304
\]

Now divide by 3:

\[
\frac{2,986,304}{3} = 995,434.67
\]

(here, exact fraction is preferable, but decimal approximation is fine for understanding).

Third Term: \(\frac{1}{2} \int_{0}^{144} (144 - x)^2 dx\)

Make the substitution:

\[
u = 144 - x \Rightarrow du = -dx
\]

When \(x=0\), \(u=144\); when \(x=144\), \(u=0\).

Reversing limits:

\[
\int{0}^{144} (144 - x)^2 dx = \int{u=144}^{0} u^2 (-du) = \int_{0}^{144} u^2 du
\]

Calculate:

\[
\int{0}^{144} u^2 du = \frac{u^3}{3} \bigg|{0}^{144} = \frac{(144)^3}{3} = 995,434.67
\]

Now multiply by \(\frac{1}{2}\):

\[
\frac{1}{2} \times 995,434.67 \approx 497,717.33
\]

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Summing It All Up

Now, combine the three parts:

\[
V = 1,492,992 - 995,434.67 + 497,717.33
\]

Calculate:

\[
V \approx (1,492,992 - 995,434.67) + 497,717.33 = 497,557.33 + 497,717.33 = 995,274.66
\]

The approximate volume of the solid is 995,275 cubic units.

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Alternative Methods and Verification

While the above approach uses direct integration, alternative methods can include:

• Changing the order of integration: Swapping the order to integrate with respect to x first.


• Using symmetry: Recognizing the symmetry of the region to simplify calculations.


• Using software