Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use cylindrical coordinates to find the indicated quantity. Volume of the solid under the surface \(z=x y\), above the \(x y\) -plane, and within the cylinder \(x^{2}+y^{2}=2 x\)

Step-by-Step Solution

Answer

The volume is \(\frac{2}{3}\).

Step 1: Convert to Cylindrical Coordinates

Cylindrical coordinates relate to Cartesian coordinates as follows: \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \). The given cylinder equation \( x^2 + y^2 = 2x \) can be rewritten in polar form. After substituting the cylindrical formulas, \( r^2 = 2r \cos \theta \), simplifying gives \( r = 2\cos\theta \).

Step 2: Set up the Integral

The volume is calculated by the integral \( V = \int \int_R z \, dA \) where \( z = xy \) which in cylindrical coordinates becomes \( z = r^2 \cos \theta \sin \theta \). The area element \( dA \) becomes \( r \, dr \, d\theta \). The limits for \( r \) are \( 0 \) to \( 2\cos\theta \) and for \( \theta \) are \( 0 \) to \( \pi/2 \).

Step 3: Evaluate the Integral

Compute the following double integral to find the volume: \[ V = \int_{0}^{\pi/2} \int_{0}^{2\cos\theta} r^3 \cos\theta \sin\theta \, dr \, d\theta \]. First, integrate with respect to \( r \): \[ \frac{r^4}{4} \bigg|_0^{2\cos\theta} = \frac{(2\cos\theta)^4}{4} \].This results in an integral with respect to \( \theta \): \[ \int_{0}^{\pi/2} \frac{16 \cos^4 \theta \cos\theta \sin\theta}{4} \, d\theta \].

Step 4: Simplify and Integrate with Respect to \(\theta\)

Simplify the integral: \[ V = 4 \int_0^{\pi/2} \cos^5\theta \sin\theta \, d\theta \]. Use the substitution method \( u = \cos\theta \), \( du = -\sin\theta \, d\theta \).This changes the limits from \(\theta = 0\) to \(\theta = \pi/2\) into \(u = 1\) to \(u = 0\), giving:\[ -4 \int_1^0 u^5 \, du = 4 \int_0^1 u^5 \, du \].

Step 5: Compute the Final Integral

Perform the final integration: \[ 4 \int_0^1 u^5 \, du = 4 \left[ \frac{u^6}{6} \right]_0^1 \]. Calculate the definite integral: \[ 4 \times \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \].

Step 6: Conclusion

The volume of the solid is \( \frac{2}{3} \). The steps involved coordinate transformation, setting up the limits, and simplifying the integral in terms of basic trigonometric identities.

Key Concepts

Volume Calculation

Understanding volume calculation in cylindrical coordinates brings efficient solutions to problems involving symmetrical solids. By transforming from Cartesian to cylindrical coordinates

• we harness the symmetry of the problem,
• simplify calculations, and
• achieve more intuitive integration limits

In this exercise, the goal is to find the volume of a solid confined by a surface above a specific region in the xy-plane. The region is bounded by the equation of a cylinder in the xy-plane. By transforming the Cartesian equation of the cylinder into polar form, we exploit the radial symmetry and set up an integral that allows the calculation of volume using cylindrical coordinates effectively.
This method is especially advantageous when dealing with rotationally symmetric objects like cylinders or spheres, where Cartesian coordinates might make calculations overly complex.

Integral Setup

Setting up the integral is a crucial step in solving volume problems using cylindrical coordinates. The integral provides the mathematical framework for accumulating an infinite number of tiny volume elements. Here:

• You convert the given surface and region into cylindrical terms.
• The function for the volume, initially stated as a product of x and y, gets converted into a function involving r, the radial distance, and \(\theta\), the angle.

The integral in polar terms becomes \( V = \int \int_R z \, dA \), where \( z = r^2 \cos \theta \sin \theta \) and \( dA \) is transformed to \( r \, dr \, d\theta \). Establishing the correct limits:

• \(r\) from 0 to \(2\cos\theta\),
• \(\theta\) from 0 to \(\pi/2\)

is essential for capturing the total volume within the cylindrical bounds. Ensuring the setup correctly defines the area of integration is key in accurately solving the volume.

Polar to Cartesian Transformation

Polar to Cartesian transformation facilitates smoother computations when dealing with cylindrical shapes. The standard Cartesian coordinates \( (x, y, z) \) are transformed to cylindrical ones using the equations:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( z = z \)

This transformation ensures the circular nature of the base is well represented, simplifying boundary conditions and integral evaluations. In this exercise, transforming the cylinder equation \( x^2 + y^2 = 2x \) into \( r^2 = 2r \cos \theta \), which simplifies further to \( r = 2 \cos \theta \), allows for straightforward calculation of the radial limits of the volume.
Assessing and implementing the transformation accurately is vital to achieve the right framework for integration.

Mathematical Integration

Mathematical integration completes the volume calculation, bridging theory and practice. Upon setting up the integral in cylindrical coordinates, performing the integration requires a systematic approach:

• Integrate with respect to \( r \) first: with the function \( r^3 \cos \theta \sin \theta \), compute \( \frac{r^4}{4} \) evaluated from 0 to \(2\cos\theta\).
• The focus then shifts to \( \theta \), resulting in an integral involving \( \cos^5\theta \sin\theta \).
• Implementing a substitution, \( u = \cos\theta \), simplifies the integral to a polynomial in \( u \).

The solution involves evaluating definite integrals with adjusted limits, leading to the final volume. In this case, \( 4 \int_0^1 u^5 \, du \) results in \( \frac{2}{3} \), offering the volume of the solid. Breaking down integration into manageable steps with clear substitutions is essential for accurate solutions.