Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the given solid. First, sketch the solid; then estimate its volume; finally, determine its exact volume. Solid between \(z=x^{2}+y^{2}+2\) and \(z=1\) and lying above \(R=\\{(x, y):-1 \leq x \leq 1,0 \leq y \leq 1\\}\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{10}{3}\) cubic units.

Step 1: Sketch the Solid

To sketch the solid, visualize the given surfaces. The surface \(z = x^2 + y^2 + 2\) is a paraboloid shifted upwards by 2 units, and \(z = 1\) is a plane parallel to the \(xy\)-plane at height 1. The region of interest, \(R\), is the rectangle defined by \(-1 \leq x \leq 1\) and \(0 \leq y \leq 1\) on the \(xy\)-plane.

Step 2: Estimate the Volume

To estimate the volume, recognize that the solid lies between the plane and the paraboloid, over the rectangle \(R\). The tallest part of the paraboloid within the rectangular region is at the point where \(x^2 + y^2\) is minimum, which is at any point on the edge where \(x = 0\) or \(y = 0\). The range of \(z\) is from the plane at 1 to some maximum \(z\)-value slightly above 1. The estimated height difference is small, indicating the volume will not be substantial.

Step 3: Setup the Volume Integral

The exact volume of the solid is given by the double integral \[ V = \iint_R ((x^2 + y^2 + 2) - 1) \, dx \, dy \].Simplify this to: \[ V = \iint_R (x^2 + y^2 + 1) \, dx \, dy \].Define \(R\) with the bounds for \(x\) and \(y\): \(-1 \leq x \leq 1\) and \(0 \leq y \leq 1\).

Step 4: Solve the Double Integral

Compute the integral: \[ V = \int_{y=0}^{1} \int_{x=-1}^{1} (x^2 + y^2 + 1) \, dx \, dy \].First, integrate with respect to \(x\):\[ \int_{x=-1}^{1} (x^2 + y^2 + 1) \, dx = \left[ \frac{x^3}{3} + (y^2+1)x \right]_{x=-1}^{1} = \frac{2}{3} + 2(y^2 + 1) = \frac{2}{3} + 2y^2 + 2 \].Now, integrate with respect to \(y\):\[ \int_{y=0}^{1} \left( \frac{2}{3} + 2y^2 + 2 \right) \, dy = \left[ \frac{2}{3}y + \frac{2}{3}y^3 + 2y \right]_{0}^{1} = \frac{2}{3} + \frac{2}{3} + 2 = \frac{10}{3} \].

Step 5: Conclusion

The exact volume of the solid is \(\frac{10}{3}\) cubic units. This volume was calculated based on the difference in height between the paraboloid and plane, integrated over the rectangular region.

Key Concepts

Double Integrals

When calculating the volume of a solid, especially one bounded by curved surfaces, double integrals come into play. Double integrals extend the idea of a single integral to calculate a "volume" under or over two-dimensional surfaces. Understanding the setup of a double integral involves visualizing how a region in the xy-plane projects onto a surface.
In practical terms, think of double integrals as adding up an infinite number of tiny volumes over a certain region. Each of these tiny volumes can be thought of as a thin rectangular column. When calculating the total volume, the double integral sums these columns together by integrating over the defined bounds for x and y.
The formula for a double integral generally looks like this:

• \([ V = \iint_R f(x, y) \, dx \, dy ]\)

where \(R\) is the region over which we're integrating. Grasping this concept is crucial for accurately determining volumes with complex surfaces.

Paraboloid

A paraboloid is a three-dimensional surface that looks like an elongated bowl. In mathematics, it's defined by the equation of the form \(z = x^2 + y^2 + c\), where "c" indicates how high the paraboloid is lifted from the xy-plane. Paraboloids are crucial in problems involving solids because they create a smooth, curved boundary.
In the exercise, the paraboloid is defined by \(z = x^2 + y^2 + 2\). This means it's a standard paraboloid shifted upward by 2 units. Paraboloids can either open upwards or downwards. The paraboloid in this instance opens upwards because as x and y increase in magnitude, z also increases.
These surfaces are frequently encountered in real-world scenarios, such as reflecting telescopes or satellites, where understanding their properties can be crucial.

Volume Calculation

Calculating the volume of a solid bounded by surfaces, such as a paraboloid, requires careful consideration of the boundaries. First, it's necessary to set the correct limits for integration, which are determined by the rectangular region of interest on the xy-plane in this case.
Volume calculation through integration involves subtracting one function from another, such as a plane from a paraboloid here. This difference represents the "height" between the two surfaces over each slice of the region \(R\). The integral sums these heights across the region to compute the entire volume:

• \([ V = \iint_R ((x^2 + y^2 + 2) - 1) \, dx \, dy ]\)

After simplifying the integrand, this equation leads to the calculation of the volume enclosed between the paraboloid and the plane.

Integration Techniques

Mastering integration techniques is essential for tackling double integrals, such as the one in this exercise. These techniques often involve multiple integration along defined axes, and understanding their properties ensures accurate solutions.
The general approach involves integrating first with respect to one variable while keeping the other fixed, and then integrating the resulting expression with respect to the second variable. For instance, in this scenario, integration with respect to x is executed first, followed by integration over y's range.
Here, you might encounter functions being integrated and anti-differentiated such as \((x^2 + y^2 + 1)\). You compute the inner integral first:

• \([ \int_{x=-1}^{1} (x^2 + y^2 + 1) \, dx ]\)

showcasing how the integration bounds influence the final result. Completing this successfully provides a well-defined volume.