At technology cyber charter school, each of the ninth grade homerooms has the same number of students and one teacher. on the last day of school, each student created a card for each of the other students in their class.
the expression 14s(s-1) can be used to find the total number of cards created by the ninth grade students. based on the given information, which of the following statements must be true? select all that apply.

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: