You are given 10cm^3 of lead that is formed into a symmetrical shape about the x

Question

You are given 10cm^3 of lead that is formed into a symmetrical shape about the x-axis. It's radius or distance from the x-axis is given by the variable y. It's length along the x-axis is 1cm. The movement of inertia for this shape, symmetric with rotation about the x-axis, is the given by the following equation,

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I= the integral of [ (1/2)(p)(pie)(y^4)dx

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Where p is the density. Find the equation for y(x) that minimizes the amount of inertia for this volume of lead. Remember that you are constrained to a fixed volume given by,

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V= the integral of [ (pie)(y^2)dx=10

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The answer is given but have no clue on how to get this answer!

Answer=> y=constant=1.8cm, making the shape a cylinder

Explanation / Answer

so we want to minimize something with a constraint so we minimize I - a(V-10) notice since V=10 this doesnt change anything in the value now we take derivative with respect to x dI/dx - a dV/dx = 0 since they are integrals with respect to x we just get back the integrand 1/2 p pi y^4 - a pi y^2 = 0 ignoring y = 0 possibility 1/2 p y^2 - a = 0 y ^2 = 2 a/p y = sqrt(2a/p) we fix a but plugging it into constraint V = integral of pi (y^2) dx = integral of pi 2 a/p dx from 0 to 1 = 2 pi a/p = 10 so 2 a/p = 10/pi plug this into y equation y= sqrt(2a/p) =sqrt(10/pi)=1.8 cm

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