Question

A solid is described along with its density function. Find the mass of the solid using cylindrical coordinates. Bounded by \(y \geq 0,\) the cylinder \(x^{2}+y^{2}=1,\) and between the planes \(z=0\) and \(z=4-y\) with density function \(\delta(x, y, z)=1\).

Short answer

The mass of the solid is \( \pi - \frac{1}{3} \).

Step-by-step solution

Convert to Cylindrical Coordinates

In cylindrical coordinates, the relations are: \( x = r\cos\theta \), \( y = r\sin\theta \), and \( z = z \). The density function given is constant \( \delta(x, y, z) = 1 \). The bounding cylinder equation \(x^2 + y^2 = 1\) translates to \(r^2 = 1\) or \( r = 1 \) in cylindrical coordinates. Thus, \( r \) ranges from 0 to 1.

Identify Bounds of Integration

The solid is bounded by \( y \geq 0 \), which implies \( \theta \) ranges from 0 to \( \frac{\pi}{2} \), encompassing the first quadrant. \( z \) is bounded below by \(z = 0\) and above by \(z = 4 - y\) which translates to \(z = 4 - r\sin\theta\) in cylindrical coordinates. Therefore, \(z\) varies from 0 to \(4 - r\sin\theta\).

Set Up the Triple Integral for Mass

The mass \( M \) of the solid is given by the triple integral of the density function over the volume of the solid:\[ M = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \int_{0}^{4-r\sin\theta} r \, dz \, dr \, d\theta \]The factor of \(r\) appears because the volume element in cylindrical coordinates is \( r \, dz \, dr \, d\theta \).

Integrate with Respect to \(z\)

First integrate with respect to \(z\):\[ \int_{0}^{4-r\sin\theta} r \, dz = r[z]_{0}^{4-r\sin\theta} = r(4-r\sin\theta) \]Simplifying, \( 4r - r^2\sin\theta \).

Integrate with Respect to \(r\)

Now integrate with respect to \(r\):\[ \int_{0}^{1} (4r - r^2\sin\theta) \, dr \]Evaluate:\[ \left[ 2r^2 - \frac{r^3\sin\theta}{3} \right]_{0}^{1} = (2 \cdot 1^2 - \frac{1^3\sin\theta}{3}) - (2 \cdot 0^2 - \frac{0^3\sin\theta}{3}) = 2 - \frac{\sin\theta}{3} \]

Integrate with Respect to \(\theta\)

Finally, integrate with respect to \(\theta\):\[ \int_{0}^{\frac{\pi}{2}} (2 - \frac{\sin\theta}{3}) \, d\theta \]Evaluate:\[ \left[ 2\theta + \frac{\cos\theta}{3} \right]_{0}^{\frac{\pi}{2}} = (2 \cdot \frac{\pi}{2} + \frac{\cos(\frac{\pi}{2})}{3}) - (2 \cdot 0 + \frac{\cos(0)}{3})\]Simplifying:\[ \pi - \frac{1}{3} \]

Conclusion

The mass of the solid is \( \pi - \frac{1}{3} \).

Key concepts

Triple Integral

A triple integral helps us calculate quantities over a three-dimensional region, such as the mass of a solid. Think of it as the process of stacking up layers within a volume to get a total value.
This is seen as an extension of double integrals, which cover areas in two dimensions. Here, each point in the volume contributes a small amount, represented by the function we are integrating (density in this problem).
The triple integral for mass of a solid is calculated as:

• Start by defining bounds for each variable (in this case, cylindrical coordinates).
• Integrate each variable step by step in sequence, updating the range for each variable as you progress.

In our example, we integrate with respect to the height, radius, and angle step-by-step. This thorough process ensures every part of the volume is accounted for.

Mass Calculation

Mass calculation involves multiplying the density of a solid by the volume over which the density is spread. When the density is constant, as in our exercise (\( \delta(x, y, z) = 1 \)), simplifying the process as you're mostly focusing on calculating the volume of the solid.
The mass of a solid is calculated using the formula:

• \[M = \int \int \int \rho(x, y, z) \ dV\\]
• This involves setting up bounds for the volume integral in the chosen coordinate system.
• We integrate the density function over these bounds using the calculated volume element.

In this exercise, the mass is simply the result of the triple integral, as the density function is constantly one.

Volume Element in Cylindrical Coordinates

In cylindrical coordinates, regions are expressed using radius (\( r \)), angle (\( \theta \)), and height (\( z \)). This transformation simplifies many problems with cylindrical symmetry.
To find the volume element (a tiny piece of volume) in cylindrical coordinates:

• We use the formula \( r \, dz \, dr \, d\theta\ \).
• This accounts for the circular nature of the 'slice' of the solid, as \( r \) scales the linear 'piece' to account for the wider circular layer.

In practice, you convert the x and y components from cartesian coordinates to \( r heta \). After transforming, the resulting integral reflects the simpler symmetry of the problem, often making it easier to solve.