Triple Integral Evaluation In Cylindrical Coordinates And Area Calculation

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This article delves into the realm of multivariable calculus, focusing on two key problems: evaluating a triple integral using cylindrical coordinates and determining the area of a region bounded by given curves. These problems showcase the power and versatility of calculus in solving geometric and analytical challenges.

(b) Evaluating the Triple Integral Using Cylindrical Coordinates

Triple integrals are powerful tools for calculating volumes and other quantities in three-dimensional space. However, evaluating them in Cartesian coordinates can sometimes be complex, especially when dealing with regions that have circular or cylindrical symmetry. This is where cylindrical coordinates come into play. Cylindrical coordinates provide a natural way to describe points in space using a radial distance, an angle, and a height, making them ideal for problems with cylindrical symmetry. The transformation from Cartesian coordinates (x, y, z) to cylindrical coordinates (r, θ, z) is given by:

  • x = r cos θ
  • y = r sin θ
  • z = z

The Jacobian determinant for this transformation is r, which means that the volume element dxdydz transforms to r dz dr dθ in cylindrical coordinates. Now, let's tackle the given integral:

∫03∫02∫04−y2dxdydz

The integral represents the volume of a region in three-dimensional space. To evaluate it using cylindrical coordinates, we first need to transform the limits of integration. The original integral has the following limits:

  • 0 ≤ z ≤ 3
  • 0 ≤ y ≤ 2
  • 0 ≤ x ≤ √(4 - y2)

The inequality 0 ≤ x ≤ √(4 - y2) represents the right half of a circle with radius 2 centered at the origin in the xy-plane. This suggests that cylindrical coordinates will be a good fit. Let's convert the limits of integration to cylindrical coordinates.

In cylindrical coordinates:

  • The equation x2 + y2 = 4 becomes r2 = 4, so r = 2.
  • The region 0 ≤ x ≤ √(4 - y2) corresponds to the right half of the circle, which means -Ï€/2 ≤ θ ≤ Ï€/2. However, since y ranges from 0 to 2, we only need the first quadrant, so 0 ≤ θ ≤ Ï€/2.
  • The limits for z remain the same: 0 ≤ z ≤ 3.
  • The limits for r are 0 ≤ r ≤ 2.

Therefore, the integral in cylindrical coordinates becomes:

∫03∫0π/2∫02rdrdθdz

Now, we can evaluate the integral step-by-step:

  1. Integrate with respect to r:

∫02rdr = [r2/2]02 = 2

  1. Integrate with respect to θ:

∫0π/22dθ = 2[θ]0π/2 = π

  1. Integrate with respect to z:

∫03πdz = π[z]03 = 3π

Thus, the value of the triple integral is 3Ï€. This represents the volume of the region, which is a quarter-cylinder with radius 2 and height 3. By converting to cylindrical coordinates, we transformed a potentially complex integral into a straightforward calculation, demonstrating the power of choosing the right coordinate system.

(c) Determining the Area of the Region

Calculating the area of a region bounded by curves is a fundamental application of integral calculus. The basic idea is to divide the region into infinitesimally small rectangles and sum their areas using an integral. When the region is bounded by curves defined by functions, we can express the area as a definite integral of the difference between the functions. Let's consider the problem of finding the area of the region bounded by the curves y = 1/x, y = √x, and the line x = 2.

To find the area of the region bounded by these curves, we first need to identify the points of intersection. These points will determine the limits of integration. Let's find the intersection points:

  1. Intersection of y = 1/x and y = √x:

Set 1/x = √x. Multiplying both sides by x, we get 1 = x√x, or x3/2 = 1. Raising both sides to the power of 2/3, we find x = 1. When x = 1, y = 1. So, the intersection point is (1, 1).

  1. Intersection of y = 1/x and x = 2:

When x = 2, y = 1/2. So, the intersection point is (2, 1/2).

  1. Intersection of y = √x and x = 2:

When x = 2, y = √2. So, the intersection point is (2, √2).

Now we know that the curves intersect at (1, 1), (2, 1/2), and (2, √2). To set up the integral, we need to determine which function is greater than the other in the region of interest. In the interval 1 ≤ x ≤ 2, √x is greater than 1/x. This can be verified by plotting the graphs of the functions or by testing a value between 1 and 2. For example, at x = 1.5, √1.5 ≈ 1.22 and 1/1.5 ≈ 0.67.

Therefore, the area of the region can be calculated as the integral of the difference between the functions y = √x and y = 1/x from x = 1 to x = 2:

Area = ∫12(√x - 1/x) dx

Now, we can evaluate the integral:

  1. Integrate √x:

∫√xdx = ∫x1/2dx = (2/3)x3/2 + C

  1. Integrate 1/x:

∫(1/x)dx = ln|x| + C

  1. Evaluate the definite integral:

Area = [(2/3)x3/2 - ln|x|]12 = [(2/3)(2)3/2 - ln(2)] - [(2/3)(1)3/2 - ln(1)]

Area = (2/3)(2√2) - ln(2) - (2/3)

Area = (4√2)/3 - ln(2) - 2/3

Thus, the area of the region bounded by the curves y = 1/x, y = √x, and the line x = 2 is (4√2)/3 - ln(2) - 2/3 square units. This result demonstrates how integral calculus can be used to find the area of complex regions by breaking them down into smaller, manageable pieces.

In Conclusion: This exploration into triple integrals and area calculations highlights the versatility and power of calculus in solving problems in geometry and other fields. Cylindrical coordinates offer a valuable tool for simplifying integrals with cylindrical symmetry, while definite integrals provide a means to calculate areas bounded by curves. These techniques are essential for anyone working with mathematical models and applications in science and engineering.

In summary, understanding and mastering these techniques are crucial for students and professionals alike. The ability to evaluate triple integrals and calculate areas accurately is a testament to one's understanding of fundamental calculus concepts and their applications. These skills are not only valuable in academic settings but also in real-world scenarios where problem-solving and analytical thinking are paramount. The knowledge gained from working through these examples serves as a strong foundation for tackling more complex problems in mathematics and related fields. This is important as it allows for a deeper appreciation of the power of calculus in solving a wide array of problems, from theoretical explorations to practical applications.