Math Problem Statement

Utilice coordenadas cilíndricas para hallar el volumen del sólido que se encuentra limitado por las superficies de ecuaciones: x^2 + y^2 + z^2 = 2z y z = sqrt(x^2 + y^2).

Solution

The problem asks to find the volume of a solid using cylindrical coordinates. The solid is bounded by the following surfaces:

  1. The equation of a sphere: x2+y2+z2=2zx^2 + y^2 + z^2 = 2z, which simplifies to (x2+y2)+(z1)2=1(x^2 + y^2) + (z - 1)^2 = 1. This describes a sphere of radius 1, centered at (0,0,1)(0, 0, 1).

  2. The surface given by z=x2+y2z = \sqrt{x^2 + y^2}, which is a cone in cylindrical coordinates.

Now, let's solve this step by step:

1. Rewrite in Cylindrical Coordinates:

In cylindrical coordinates, the transformations are: x=rcos(θ),y=rsin(θ),z=zx = r \cos(\theta), \quad y = r \sin(\theta), \quad z = z where r=x2+y2r = \sqrt{x^2 + y^2}.

The equations become:

  • For the sphere: (x2+y2)+(z1)2=1(x^2 + y^2) + (z - 1)^2 = 1 becomes r2+(z1)2=1r^2 + (z - 1)^2 = 1.
  • For the cone: z=x2+y2=rz = \sqrt{x^2 + y^2} = r.

2. Find the Region of Integration:

From the sphere equation r2+(z1)2=1r^2 + (z - 1)^2 = 1, we solve for zz: z=1±1r2z = 1 \pm \sqrt{1 - r^2} Thus, the upper boundary of zz is z=1+1r2z = 1 + \sqrt{1 - r^2} and the lower boundary is z=11r2z = 1 - \sqrt{1 - r^2}, but we are only interested in the portion below the cone, where z=rz = r.

3. Volume Integral in Cylindrical Coordinates:

The volume integral in cylindrical coordinates is given by: V=θ=02πr=01z=r1+1r2rdzdrdθV = \int_{\theta=0}^{2\pi} \int_{r=0}^{1} \int_{z=r}^{1 + \sqrt{1 - r^2}} r \, dz \, dr \, d\theta Here, rdzdrdθr \, dz \, dr \, d\theta is the volume element in cylindrical coordinates.

4. Compute the Volume:

Let's proceed to compute this integral step by step. I will calculate it for you.

The volume of the solid bounded by the given surfaces, computed using cylindrical coordinates, is π\pi cubic units.

Would you like more details on the steps or any specific clarifications?

Here are five related questions to expand on the topic:

  1. How do cylindrical coordinates simplify volume integrals in problems with symmetry?
  2. How would the integral change if the solid had different boundary surfaces?
  3. Can this problem be solved using spherical coordinates, and how would the approach differ?
  4. How is the Jacobian determinant used when converting from Cartesian to cylindrical coordinates?
  5. What are other common examples of solids where cylindrical coordinates are useful?

Tip: When solving volume problems with symmetry, choosing the appropriate coordinate system (like cylindrical or spherical) often simplifies the integral significantly.

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Math Problem Analysis

Mathematical Concepts

Cylindrical Coordinates
Volume of Solids
Multivariable Calculus

Formulas

x = r cos(θ), y = r sin(θ), z = z
Volume integral in cylindrical coordinates: V = ∫∫∫ r dz dr dθ
Equation of a sphere: r^2 + (z - 1)^2 = 1
Equation of a cone: z = r

Theorems

Cylindrical Coordinate System
Integration in Three Dimensions

Suitable Grade Level

Undergraduate - Calculus III