Find the mass of the two-dimensional object. An oversized hockey puck of radius  
4
  in. with a density function of  
ρ
(
x
)
=
x
3
−
5
x
+
5
 .

To find the mass of the two-dimensional object (the hockey puck), we need to integrate the density function over the area of the puck. The mass of a two-dimensional object is given by the integral of the density function over the region.

Given:
- Radius \( r = 4 \) inches
- Density function \( \rho(x) = x^3 - 5x + 5 \)

Since this is a circular region with radius \( r = 4 \), the mass of the puck is calculated using polar coordinates due to its symmetry. However, since the density function \( \rho(x) \) depends only on the radial distance \( x \), we will integrate over the area of the puck.

### Step 1: Set up the integral
The formula for mass is:

\[
M = \int_{0}^{r} \int_{0}^{2\pi} \rho(r) \, r \, d\theta \, dr
\]

In our case, the density \( \rho(x) = x^3 - 5x + 5 \) is a function of \( r \), the radial distance. So we need to replace \( \rho(r) \) with \( r^3 - 5r + 5 \).

Thus, the mass integral becomes:

\[
M = \int_0^4 \int_0^{2\pi} (r^3 - 5r + 5) r \, d\theta \, dr
\]

### Step 2: Perform the angular integration
The angular integral over \( \theta \) from 0 to \( 2\pi \) simplifies:

\[
\int_0^{2\pi} d\theta = 2\pi
\]

So the mass integral reduces to:

\[
M = 2\pi \int_0^4 (r^4 - 5r^2 + 5r) \, dr
\]

### Step 3: Integrate with respect to \( r \)
Now we integrate the function \( r^4 - 5r^2 + 5r \) over \( r \) from 0 to 4:

\[
\int_0^4 (r^4 - 5r^2 + 5r) \, dr = \left[ \frac{r^5}{5} - \frac{5r^3}{3} + \frac{5r^2}{2} \right]_0^4
\]

Evaluating this at the limits:

\[
= \left( \frac{4^5}{5} - \frac{5 \times 4^3}{3} + \frac{5 \times 4^2}{2} \right) - \left( 0 \right)
\]

\[
= \left( \frac{1024}{5} - \frac{5 \times 64}{3} + \frac{5 \times 16}{2} \right)
\]

Simplify:

\[
= \left( 204.8 - \frac{320}{3} + 40 \right)
\]

Now, calculate each term:

\[
204.8 - 106.67 + 40 = 138.13
\]

### Step 4: Multiply by \( 2\pi \)
Finally, multiply by \( 2\pi \) to get the mass:

\[
M = 2\pi \times 138.13 = 276.26\pi \, \text{in}^2
\]

Thus, the mass of the hockey puck is approximately:

\[
M \approx 867.97 \, \

Find the mass of the two-dimensional object. An oversized hockey puck of radius 4 inches with a density function ρ(x) = x^3 - 5x + 5.

Learn how to find the mass of a two-dimensional hockey puck with a radius of 4 inches and a density function ρ(x) = x^3 - 5x + 5. This detailed solution walks through using polar coordinates and integration techniques, perfect for college-level calculus students.

Math Problem Statement

Find the mass of the two-dimensional object. An oversized hockey puck of radius
4 in. with a density function of
ρ ( x )

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