Find the surface area of the frustum of a cone described as the portion of the cone z =...

Question:

Find the surface area of the frustum of a cone described as the portion of the cone z = {eq}\sqrt{x^2+y^2} {/eq}that is between z = 1 and z = 2.

Surface Area of frustum:

To find the surface area of the frustum, first find the partial differentiation of the given function with respect to x and y.

After that, substitute the derivatives in the formula.

The formula is: {eq}\int \int _{D} \sqrt{1 + \left ( \frac{\partial z}{\partial x} \right )^{2} + \left ( \frac{\partial z}{\partial y} \right )^{2}} dy dx {/eq}

Now, solve the integrals.

Answer and Explanation:

Given: {eq}z = \sqrt{x^{2} + y^{2}} {/eq}

{eq}z = 1 \: and \: z = 2 {/eq}

Take partial differentiation of the given function with respect to x and y.

{eq}\frac{\partial z}{\partial x} = \frac{x}{\sqrt{x^{2} + y^{2}}} {/eq}

{eq}\frac{\partial z}{\partial y} = \frac{y}{\sqrt{x^{2} + y^{2}}} {/eq}

Find the area of frustum.

{eq}\int \int _{D} \sqrt{1 + \left ( \frac{\partial z}{\partial x} \right )^{2} + \left ( \frac{\partial z}{\partial y} \right )^{2}} dy dx = \int \int _{D} \sqrt{1 + \left ( \frac{x}{\sqrt{x^{2} + y^{2}}} \right )^{2} + \left ( \frac{x}{\sqrt{x^{2} + y^{2}}} \right )^{2}} dy dx {/eq}

{eq}= \int \int _{D} \sqrt{1 + \frac{x^{2}}{x^{2} + y^{2}} + \frac{y^{2}}{x^{2} + y^{2}}} dy dx {/eq}

{eq}= \int \int _{D} \sqrt{\frac{x^{2} + y^{2} + x^{2} + y^{2}}{x^{2} + y^{2}} } dy dx {/eq}

{eq}=\int \int _{D} \sqrt{2 } dy dx {/eq}

Change the Cartesian coordinate into polar coordinates.

{eq}\int \int _{D} \sqrt{1 + \left ( \frac{\partial z}{\partial x} \right )^{2} + \left ( \frac{\partial z}{\partial y} \right )^{2}} dy dx = \sqrt{2} \int_{0}^{2\pi } \int_{1}^{2} r dr d\theta {/eq}

{eq}= \sqrt{2} \left [\frac{r^{2}}{2} \right ]_{1}^{2} \int_{0}^{2\pi } d\theta {/eq}

{eq}= \sqrt{2} \left [\frac{2^{2} - 1^{2}}{2} \right ] \left [ \theta \right ]_{0}^{2\pi } {/eq}

{eq}= \sqrt{2} \left [\frac{3}{2} \right ]2\pi {/eq}

{eq}= 3\sqrt{2}\pi {/eq}

Hence, the surface area is {eq}3\sqrt{2}\pi {/eq}.


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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