Find the mass of a thin funnel in the shape of a cone $z = \sqrt{x^2 + y^2}$, $1 \leq z \leq 3$, if its density function is $\rho(x, y, z) = 5 - z$.

Neetesh Kumar | November 13, 2024

Calculus Homework Help

This is the solution to Math 1D
Assignment: 16.7 Question Number 20
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Step-by-step solution:

To find the mass of the funnel, we use the surface integral:

$\text{Mass} = \iint_{S} \rho \, dS$

where $\rho(x, y, z) = 5 - z$ is the density function, and $dS$ represents the surface element of the cone.

Step 1: Parametrize the Surface

The surface of the cone is given by $z = \sqrt{x^2 + y^2}$, which implies that $r = \sqrt{x^2 + y^2} = z$ in cylindrical coordinates. Thus, the surface can be parametrized as:

$x = r \cos \theta$, $y = r \sin \theta$, and $z = r$, with $1 \leq r \leq 3$.

Step 2: Compute $dS$

The surface element $dS$ for a surface parametrized by $z = f(x, y)$ is:

$dS = \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dx \, dy$

Since $z = f(x, y) = \sqrt{x^2 + y^2}$, we compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

1. $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{z}$
2. $\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{y}{z}$

Then,

$dS = \sqrt{1 + \left( \frac{x}{z} \right)^2 + \left( \frac{y}{z} \right)^2} \, dx \, dy$

Simplify the expression under the square root:

$dS = \sqrt{1 + \frac{x^2 + y^2}{z^2}} \, dx \, dy$

Since $x^2 + y^2 = z^2$ (from the cone equation), this becomes:

$dS = \sqrt{1 + \frac{z^2}{z^2}} \, dx \, dy = \sqrt{2} \, dx \, dy$

Thus, $dS = \sqrt{2} \, r \, dr \, d\theta$ in cylindrical coordinates.

Step 3: Set Up the Integral

The mass of the funnel is:

$\text{Mass} = \iint_{S} \rho \, dS = \int_{0}^{2\pi} \int_{1}^{3} (5 - r) \sqrt{2} \, r \, dr \, d\theta$

Factor out $\sqrt{2}$:

$= \sqrt{2} \int_{0}^{2\pi} d\theta \int_{1}^{3} (5 - r) \, r \, dr$

Separate the integrals:

$= \sqrt{2} \int_{0}^{2\pi} d\theta \int_{1}^{3} (5r - r^2) \, dr$

Step 4: Evaluate Each Integral

1. $\int_{0}^{2\pi} d\theta = 2\pi$
2. $\int_{1}^{3} (5r - r^2) \, dr = \left[ \frac{5r^2}{2} - \frac{r^3}{3} \right]_{1}^{3}$

Evaluate the second integral:

$\int_{1}^{3} (5r - r^2) \, dr = \left( \frac{5 \cdot 3^2}{2} - \frac{3^3}{3} \right) - \left( \frac{5 \cdot 1^2}{2} - \frac{1^3}{3} \right)$

$= \left( \frac{45}{2} - 9 \right) - \left( \frac{5}{2} - \frac{1}{3} \right)$

Convert terms to a common denominator where necessary:

$= \left( \frac{45}{2} - \frac{18}{2} \right) - \left( \frac{15}{6} - \frac{2}{6} \right)$

$= \frac{27}{2} - \frac{13}{6} = \frac{81 - 13}{6} = \frac{68}{6} = \frac{34}{3}$

Now, combine results:

$\text{Mass} = \sqrt{2} \cdot 2\pi \cdot \frac{34}{3}$

$= \frac{68\pi \sqrt{2}}{3}$

Final Answer

The mass of the funnel is $\boxed{\dfrac{68\pi \sqrt{2}}{3}}$.

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