Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

The part of the surface \(z=x^{2} / 4+4\) that is cut off by the planes \(x=0, x=1, y=0\), and \(y=2\)

Step-by-Step Solution

Answer

The area of the cut-off surface is \( \frac{49}{6} \).

Step 1: Recognize the Surface Equation

The surface given is represented by the equation \( z = \frac{x^2}{4} + 4 \). This is a parabolic cylinder in three-dimensional space, opening upwards along the \( z \)-axis.

Step 2: Identify the Bounding Planes

The planes mentioned in the problem are \( x = 0 \), \( x = 1 \), \( y = 0 \), and \( y = 2 \). These planes form a rectangular boundary in the \( xy \)-plane.

Step 3: Determine the Bounds of Integration

Given the planes, the integration limits for \( x \) are from 0 to 1, and for \( y \) are from 0 to 2. This frames a rectangular region on the \( xy \)-plane.

Step 4: Setup the Double Integral

The area of the rectangular region beneath the surface can be calculated by integrating the surface equation over the specified bounds. Set up the double integral: \[ \int_{0}^{2} \int_{0}^{1} \left( \frac{x^2}{4} + 4 \right) \, dx \, dy \]

Step 5: Evaluate the Inner Integral

Evaluate the integral with respect to \( x \):\[ \int_{0}^{1} \left( \frac{x^2}{4} + 4 \right) \, dx = \left[ \frac{x^3}{12} + 4x \right]_{0}^{1} = \frac{1^3}{12} + 4(1) = \frac{1}{12} + 4 = \frac{49}{12} \]

Step 6: Evaluate the Outer Integral

Substitute the result from the inner integral and integrate with respect to \( y \):\[ \int_{0}^{2} \frac{49}{12} \, dy = \frac{49}{12} \times y \Big|_{0}^{2} = \frac{49}{12} \times 2 = \frac{98}{12} = \frac{49}{6} \]

Step 7: Conclusion

The area of the part of the surface \( z = \frac{x^2}{4} + 4 \) cut off by the planes is \( \frac{49}{6} \). Thus, the volume above the rectangular region and below the surface is \( \frac{49}{6} \).

Key Concepts

Surface Equation

When dealing with a surface equation in calculus, it helps to visualize what the equation represents in a three-dimensional space. For the exercise at hand, the surface is described by the equation:
\[ z = \frac{x^2}{4} + 4 \]
This is a specific type of surface known as a parabolic cylinder. The equation describes how high the surface is (the value of \( z \)) for each point \((x, y)\) on the surface. Think of it as mapping each point in the \((x, y)\)-plane to a height \( z \). By understanding the surface equation, you can predict the shape formed in 3D space which "opens" or curves upward parallel to the \( z \)-axis. Since the surface is defined by the square of \( x \) divided by 4, plus 4, each parallel slice (for constant \( y \)) exhibits the shape of a parabola.

Parabolic Cylinder

A parabolic cylinder is a 3D shape generated by extending a parabola along an axis, in this case, the y-axis. To visualize:

• The base shape is a parabola (like \( y = \frac{x^2}{4} \)) in the xz-plane.
• This is then extended along the y-axis, forming a cylindrical shape that curves upward.

In the context of our surface equation, every slice parallel to the yz-plane remains parabolic as it travels along the y-axis. This results in a shape that looks similar to a "slide," with the bottom curving upwards and outwards. Understanding parabolic cylinders makes it easier to conceive how the surface 'sits' in space and helps in calculating volumes or areas concerning these surfaces.

Bounds of Integration

The concept of integration bounds is crucial when performing any calculus operation over a region. In this problem, the region of interest is a rectangular zone defined by the intersection of presented planes:

• \(x = 0\) and \(x = 1\) mark the limits for the x-axis.
• \(y = 0\) and \(y = 2\) set the limits for the y-axis.

The cross-section of these planes in the xy-plane forms a rectangle. Understanding these boundaries is crucial because they define our "area of interest" for the double integral. These planes "cut off" a section of the parabolic cylinder — meaning only the portion contained within these bounds should be considered when calculating volume.

Volume Calculation

Calculating volume under a surface over a defined region involves evaluating a double integral. The double integral sums up all infinitesimally small "columns" of volume beneath the surface:
\[\int_{0}^{2} \int_{0}^{1} \left( \frac{x^2}{4} + 4 \right) \, dx \, dy\]
This integral shows that we first integrate with respect to \( x \) — dealing with the volume in one direction, and then with respect to \( y \) — adding it up across the second direction. In the solution:

• The inner integral evaluates how much height is captured by moving from \( x=0 \) to \( x=1 \).
• Once this is calculated, the outer integral completes the volume sum by scanning from \( y=0 \) to \( y=2 \).

After integration, the volume calculated is \( \frac{49}{6} \). This is the '3D measure' of the cut-off section of the surface. Understanding these steps is key to comprehending how calculus can effectively quantify space under complex surfaces.