How to find moment of inertia of lemniscate
r^2=a^2cos2@ about z axis
me

introduction

we have here a lemniscate[1] with the equation in polar form as,

r2=a2cos2θ(1)

the graph of which at a=1 looks like:

pretty.

setting up coordinate and integration limits

i will consider the polar coordinates. therefore let’s clear out some basic transformations.

x=rcosθ

y=rsinθ

da=rdrdθ

moving on i shall set the limit of integration. pretty obviously r shall go from 0 to acos2θ√. for θ however we need to define a domain where the area will converge. setting r=0, we shall get from equation 1, θ=π4+nπ2 where n∈z. now from the graph we can see that the function is even, and the area converges between θ from −π/4 to π/4 to get covering one side. therefore we can integrate θ from −π/4 to π/4 to get one half and then multiply by 2.

calculations for the area

we need the area to calculate the density. therefore we have,

a=2∫π/4θ=−π/4∫acos2θ√r=0rdrdθ

=∫π/4−π/4a2cos2θdθ

=a2sin2θ2∣∣∣π/4−π/4

a=a2

calculation for ixx

since i don’t know which axis you specifically want, i will calculate over all the axis as well as the production of inertia and then get its inertia tensor.

i have taking density as σ,

ixx=2σ∫π/4θ=−π/4∫acos2θ√r=0r2sin2θrdrdθ

=σ2∫π/4−π/4a4cos22θsin2θdθ

=σ∫π/40a4cos22θsin2θdθ

=σ2∫π/40a4cos22θ(1−cos2θ)dθ

=σ2{∫π/40a4cos22θdθ−∫π/40a4cos32θdθ}

putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,

=σa44{∫π/20cos2ϕdϕ−∫π/20cos3ϕdϕ}

now using beta function[2] we have here, p=1/2 and q=3/2 for the first integral and p=1/2 and q=2 for the second integral. therefore we get using gamma-beta relation,

=σa44{π4−23}

expanding σ,

ixx=ma248(3π−8)

calculation for iyy

iyy=2σ∫π/4θ=−π/4∫acos2θ√r=0r2cos2θrdrdθ

=σ2∫π/4−π/4a4cos22θcos2θdθ

=σ∫π/40a4cos22θcos2θdθ

=σ2∫π/40a4cos22θ(cos2θ+1)dθ

=σ2{∫π/40a4cos32θdθ+∫π/40a4cos22θdθ}

putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,

=σa44{∫π/20cos3ϕdϕ+∫π/20cos2ϕdϕ}

now using beta function we have here, p=1/2 and q=2 for the first integral and p=1/2 and q=3/2 for the second integral. therefore we get using gamma-beta relation,

=σa44{23+π4}

expanding σ,

iyy=ma248(8+3π)

calculation for izz

since the object considered is 2d we can use the perpendicular axis theorem to get izz. we have,

izz=ixx+iyy

izz=mπa28

calculation for ixy and iyx

moving on i shall calculate products of inertia. we have here,

ixy=iyx=2σ∫π/4θ=−π/4∫acos2θ√r=0r2sinθcosθrdrdθ

=σ∫π/4−π/4∫acos2θ√0r2sin2θrdrdθ

=σ4∫π/4−π/4a4cos22θsin2θdθ

=σ2∫π/40a4cos22θsin2θdθ

putting ϕ=2θ and dϕ=2dθ, we will have limits, ϕ from 0 to π/2. therefore,

=σa44∫π/20cos2ϕsinϕdϕ

now using beta function we have here, p=1 and q=3/2 for the first integral and p=1/2 and q=3/2 for the second integral. therefore we get using gamma-beta relation,

=σa412

expanding density,

ixy=iyx=ma212

calculation for ixz and izx

since the object is 2d, z=0, hence,

ixz=izx=0

calculation for iyz and izy

again since the object is 2d, z=0, hence,

iyz=izy=0

inertia tensor of a lemniscate

finally as promised, here is your inertia tensor:

i↔=⎡⎣⎢⎢⎢ma248(3π−8)ma2120ma212ma248(3π+8)000mπa28⎤⎦⎥⎥⎥

i was thinking of calculating the ellipsoid of inertia as well, however it’s quite late night and i am kind of sleepy, so maybe later. also since these calculations were pesky if you do find any mistake, comment it down or suggest an edit.

step-by-step explanation:

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.

answer: 30

step-by-step explanation: its 30 because im a basketball player and stephen curry hits most of his shots so if you put 30 you might make the shot goodluck