Summary
Keywords
Full Transcript
We look at triple integrals of w=f(x,y,z) over regions that are best described with cylindrical (polar) coordinates. This is Multivariable Calculus Unit 5 Lecture 3. Cylindrical coordinates are an extension of polar coordinates to three dimensions, ideal for describing shapes whose shadows in the 𝑥𝑦-plane are round. The coordinates are defined as 𝑥=𝑟cos(𝜃), 𝑦=𝑟sin(𝜃), and 𝑧=𝑧 (no conversion). Note that the 𝑟 and 𝜃 components lie in the 𝑥𝑦-plane and are not the displacement from the point to the origin in 3D space. The process of setting up a triple integral in cylindrical coordinates is similar to double integration in polar coordinates. The primary difference is the inclusion of the 𝑧-coordinate, which remains unchanged. When converting the integral, 𝑑𝑥𝑑𝑦 is replaced with 𝑟𝑑𝑟𝑑𝜃, and 𝑑𝑧 remains as is. Example 1: Integrating over a Solid Cone. Here's how the given bounds can be converted to cylindrical coordinates: 1. Bounds for 𝑦: The middle integral limits are from −sqrt(9−𝑥^2) to sqrt(9−𝑥^2), which are the bounds on 𝑦 for a circle of radius 3 centered at the origin in the 𝑥𝑦-plane. This is because the equations 𝑦=±sqrt(9−𝑥^2) describe the upper and lower semicircles of a circle with radius 3. 2. Bounds for 𝑥: The outer integral limits are from −3 to 3, which combined with the first analysis suggests that 𝑥 is taking all values across the diameter of a circle centered at the origin in the 𝑥𝑦-plane, with a radius of 3. 3. Bounds for 𝑧: The inner integral limits are from sqrt(𝑥^2 + 𝑦^2) to 3. This describes a region starting from the cone 𝑧 = sqrt(𝑥^2 + 𝑦^2) and extending up to the plane 𝑧=3. In cylindrical coordinates, we have the following substitutions and interpretations: 𝑥=𝑟cos(𝜃) 𝑦=𝑟sin(𝜃) 𝑧 remains 𝑧 With these substitutions: The bounds for 𝑟 would be from 0 (at the center of the circle) to 3 (the radius of the circle). The bounds for 𝜃 would be from 0 to 2𝜋 to cover the entire circle. The bounds for 𝑧 become from 𝑟 (the cone) to 3 (the plane). Example 2: Integrating over a Quarter Cylinder. Here's how to convert the limits to cylindrical coordinates: 1. Bounds for 𝑦: The 𝑦 limits go from 0 to sqrt(1-x^2), which are the bounds for 𝑦 between the 𝑥-axis and the upper semicircle or radius 1. 2. Bounds for 𝑥: The 𝑥 limits go from 0 to 1, indicating that we want the solid semidisk in the first quadrant of the 𝑥𝑦-plane. 3. Bounds for 𝑧: The 𝑧 limits go from 0 to 4, which suggests 𝑧 is free to move up to a height of 4 above the xy-plane. In cylindrical coordinates, we make the following substitutions: 𝑥=𝑟cos(𝜃) 𝑦=𝑟sin(𝜃) 𝑧 remains 𝑧 With these substitutions: The bounds for 𝑟 would be from 0 (at the center of the circle) to 1 (the radius of the circle in the 𝑥𝑦-plane). The bounds for 𝜃 would be from 0 to 𝜋2 to cover the first quadrant of the circle. The bounds for 𝑧 remain from 0 to 4. Example 3: Integrating over a Region between a Cone and a Sphere. To convert the description of the volume of the solid region above the cone 𝑧 = sqrt(𝑥^2+𝑦^2) and below the sphere 𝑥^2+𝑦^2+𝑧^2=8 in the first octant to cylindrical coordinates for a triple integral, we proceed as follows: 1. The Cone 𝑧 = sqrt(𝑥^2+𝑦^2): In cylindrical coordinates, this cone is simply described by 𝑧 = 𝑟, because 𝑥^2+𝑦^2 = 𝑟^2 by definition. 2. The Sphere 𝑥^2+𝑦^2+𝑧^2=8: In cylindrical coordinates, this sphere becomes 𝑟^2+𝑧^2=8. To get 𝑧 by itself, we take the square root of both sides, giving 𝑧=sqrt(8-r^2), which describes the upper half of the sphere. 3. The First Octant: The first octant is where 𝑥,𝑦, and 𝑧 are all non-negative. In cylindrical coordinates, this translates to 𝑟 ≥ 0, 𝜃 ranging from 0 to 𝜋/2, and 𝑧 ≥ 0. Putting this together, the bounds for the triple integral in cylindrical coordinates are: 𝜃 bounds: Since we're in the first octant, 𝜃 ranges from 0 to 𝜋/2. 𝑟 bounds: 𝑟 starts at 0 at the vertex of the cone and extends out to the intersection with the sphere. The intersection occurs where 𝑟^2+𝑟^2=8, so 𝑟^2=4 or 𝑟=2. Hence, 𝑟 ranges from 0 to 2. 𝑧 bounds: For a given 𝑟 in the region, 𝑧 starts at the cone 𝑧=𝑟 and goes up to the sphere 𝑧=sqrt(8-r^2). #multivariablecalculus #mathematics #iitjammathematics #calculus3 #TripleIntegration #VolumeCalculation #MathLecture #Calculus3 #IntegrationTechniques #MathTutorial
Continue this lesson in the app
Install CourseHive on Android or iOS to keep learning while you move.
