Find the moment of inertia about the z - axis of a thin shell of constant density \delta cut...

Question:

Find the moment of inertia about the z - axis of a thin shell of constant density {eq}\delta {/eq} cut from the cone {eq}4x^2 + 4y^2 - z^2 = 0; \quad z \geq 0 {/eq} by the circular cylinder {eq}x^2 + y^2 = 2x {/eq}

Double Integral

This problem is based on the application of the double integral. We are going to use the concept of finding the moment of inertia using integral as well as the concept of polar coordinates to simplify our integrals.

Answer and Explanation:


We are aiming to find the moment of inertia about the z-axis.

It can be given as,

{eq}\displaystyle I_z=\int (x^2+y^2)dm {/eq}

Where, {eq}dm {/eq} is the infinitesimal mass.

{eq}\displaystyle I_z=\int \int (x^2+y^2)\delta ds {/eq}, where, {eq}\delta {/eq} is the density.

{eq}\displaystyle I_z=\delta \int \int (x^2+y^2)\sqrt{1+z^2_x+z^2_y}dA {/eq}

Now,

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

{eq}\displaystyle z_x=\frac{2x}{\sqrt{x^2+y^2}} {/eq}

{eq}\displaystyle z_y=\frac{2y}{\sqrt{x^2+y^2}} {/eq}


Plugging in the values, we get,

{eq}\displaystyle I_z=\delta \int \int (x^2+y^2)\sqrt{1+\left ( \frac{4x^2}{x^2+y^2} \right )+\left (\frac{4y^2}{x^2+y^2} \right )}dA {/eq}

Using polar coordinates, we get,

{eq}\displaystyle I_z=\delta\sqrt 5 \int \int (x^2+y^2)dA {/eq}

{eq}\displaystyle I_z=\delta\sqrt 5 \int_{0}^{\pi}\int_{0}^{2\cos\theta} r^2.r\ drd\theta {/eq}

{eq}\displaystyle I_z=4\sqrt 5 \delta \int_{0}^{\pi}\ \cos^4 \theta d\theta {/eq}

{eq}\displaystyle I_z=4\sqrt 5 \delta \left ( \frac{3\pi}{8} \right ) {/eq}

Thus, the moment of inertia about the z-axis is:

{eq}\displaystyle \boxed{\displaystyle I_z=\frac{(3\sqrt 5)\pi \delta}{2}} {/eq}


Learn more about this topic:

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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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