Question

By transforming to cylindrical polar coordinates, evaluate the integral $$ I=\iiint \ln \left(x^{2}+y^{2}\right) d x d y d z $$ over the interior of the conical region \(x^{2}+y^{2} \leq z^{2}, 0 \leq z \leq 1\)

Short answer

- (5\pi)/9

Step-by-step solution

Transform the coordinates

Convert Cartesian coordinates to cylindrical coordinates. Use the relations: \[ x = r \, \cos(\theta), \quad y = r \, \sin(\theta), \quad z = z \] Also, the volume element transforms as: \[ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz \]

Set up the integral in cylindrical coordinates

In cylindrical coordinates, the integral becomes: \[ I = \iiint \, \ln(r^2) \, r \, dr \, d\theta \, dz \]

Determine the limits of integration

Identify the region of integration. Given: \[ 0 \le z \le 1 \] and \[ 0 \le r \le z \] and \[ 0 \le \theta < 2\pi \]

Evaluate the integral over \( \theta \)

Integrate with respect to \( \theta \) first: \[ \int_0^{2\pi} \, d\theta = 2\pi \]

Evaluate the integral over \( z \)

Integrate with respect to \( z \): \[ I = 2\pi \int_0^1 \left( \int_0^z \, r \, \ln(r^2) \, dr \right) dz \]

Simplify and integrate over \( r \)

Solve the inner integral: \[ \int_0^z \, \ln(r^2) \, r \, dr = \left[ \frac{1}{2} \, r^2 \, \ln(r^2) \right]_0^z - \int_0^z r \, dr \] \[ = \frac{1}{2} \, z^2 \, \ln(z^2) - \left[ \frac{1}{2} r^2 \right]_0^z \] \[ = \frac{1}{2} \, z^2 \, (2 \ln(z)) - \frac{1}{2} \, z^2 \] \[ = z^2 \, \ln(z) - \frac{1}{2} \, z^2 \]

Final integration over \( z \)

Integrate the result with respect to \( z \): \[ I = 2\pi \int_0^1 \left( z^2 \, \ln(z) - \frac{1}{2} \, z^2 \right) dz \int_0^1 \, z^2 \, \ln(z) \, dz \] Use integration by parts for the first term and standard integration for the second term.

Compute integrals

For the term involving \( \ln(z) \): Integration by parts: Let \[ u = \ln(z) \quad dv = z^2 \, dz \] Then, \[ du = \frac{1}{z} dz \quad v = \frac{z^3}{3} \] \[ \int_0^1 z^2 \ln(z) \ dz = \left[ \frac{z^3}{3} \ln(z) \right]_0^1 - \int_0^1 \frac{z^3}{3} \, \frac{1}{z} \, dz \] \[ = \left( \frac{1}{3} \ln(1) - \left( 0 \right) \right) - \frac{1}{3} \int_0^1 z^2 \, dz \] Remaining integral: \[ - \frac{1}{3} \int_0^1 z^2 dz = - \frac{1}{3} \left[ \frac{z^3}{3} \right]_0^1 \] \[ = - \frac{1}{3} \times \frac{1}{3} = - \frac{1}{9} \] Adding the integral of \( \frac{-z^2}{2} \): \[ \int_0^1 - \frac{1}{2} z^2 dz = \frac{-1}{2} \left[ \frac{z^3}{3} \right]_0^1 = - \frac{1}{6} \] Summing both results: \[ 2 \pi \left( - \frac{1}{9} - \frac{1}{6} \right) = 2 \pi \times - \frac{5}{18} = - \frac{5 \pi}{9} \]

Key concepts

Cylindrical Coordinates

In mathematics and physics, cylindrical coordinates are a system of coordinates that extend the two-dimensional polar coordinates to three dimensions by using an additional height (z) coordinate. This system is particularly advantageous when dealing with problems exhibiting cylindrical symmetry, such as evaluating integrals over certain regions. The relations for converting Cartesian coordinates \(x, y, z\) to cylindrical coordinates \(r, \theta, z\) are given by: \[ x = r \cos(\theta), \quad y = r \sin(\theta), \quad z = z \] Furthermore, the volume element \(dV\) in cylindrical coordinates transforms into: \[ dV = dx \, dy \, dz = r \, dr \, d\theta \, dz \] This transformation simplifies the integration process for regions with cylindrical symmetry, making it easier to handle \(r\) and \(\theta\) components.
This transformation is essential in the given problem as it simplifies the given integral.

Volume Integral

A volume integral extends the concept of integrating a function over a region into three dimensions. It involves calculating the integral of a function \(f(x, y, z)\) over a volume \(V\) in space. In cylindrical coordinates, the volume integral can be expressed as: \[ \iiint_V f(r, \theta, z) \, r \, dr \, d\theta \, dz \] The given problem requires evaluating a volume integral by transforming to cylindrical coordinates. The integral: \[ I = \iiint_V \ln(x^2 + y^2) \, dx \, dy \, dz \] becomes: \[ I = \iiint_V \, \ln(r^2) \, r \, dr \, d\theta \, dz \] The function \(\ln(x^2 + y^2)\) simplifies to \(\ln(r^2)\) in cylindrical coordinates due to the relationship between \(x, y\) and \(r\).
The volume element \(r \, dr \, d\theta \, dz\) represents the small volume in cylindrical coordinates, facilitating easier integration over the conical region.

Integration by Parts

Integration by parts is a mathematical technique used to integrate products of functions. It is based on the product rule for differentiation and is particularly useful when direct integration is challenging. The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \] where you choose \(u\) and \(dv\) such that \(du\) and \(v\) are easily computable.
In the problem, we use integration by parts for the term involving \(\ln(z)\):
Let \u = \ln(z)\ and \dv = z^2 \, dz\. Then, \du = \frac{1}{z} dz\ and \v = \frac{z^3}{3}\.
When evaluating: \[ \int_0^1 z^2 \, \ln(z) \, dz = \left[ \frac{z^3}{3} \ln(z) \right]_0^1 - \int_0^1 \frac{z^3}{3} \, \frac{1}{z} \, dz \] We solve the integrals separately to find the final value, accounting for boundary conditions. This process simplifies complex integrals by breaking them into more manageable parts.

Triple Integral

A triple integral extends the concept of double integrals to three dimensions. It involves integrating a function \(f(x, y, z)\) over a three-dimensional region. In cylindrical coordinates, a triple integral can be represented as: \[ \iiint_V f(r, \theta, z) \, r \, dr \, d\theta \, dz \] This approach is an effective way to compute the volume under surfaces or find mass, charge, and other physical properties over a region.
For our problem, the triple integral evaluates: \[ I = 2\pi \int_0^1 \left( z^2 \, \ln(z) - \frac{1}{2} \, z^2 \right) dz \] The limits of integration \(0 \leq z \leq 1\), \(0 \leq r \leq z\), and \(0 \leq \theta < 2\pi\) define the conical region's bounds.
Breaking down the integration into \(\theta\), \(r\), and \(z\) components streamlines evaluating the integral. By transforming to cylindrical coordinates and using techniques such as integration by parts, we efficiently resolve the complex volume integral over the specified region.