Recent content by Lucas Mayr

Ok, so I've tried looking at dT as a function of both P and v and reached dT = (∂T/∂P)v dP + (∂T/∂v)P dv And after reducing the derivatives, dT = KT/α dP + 1/vα dv , and using the problem's KT and α. dT = 1/D dv + Ev2/D dP dT = 1/D dv + EPava2/(PbD) dP Tb = Ta + (vb - va)/D + EPava2 Ln(Pb/Pa)/D...

Homework Statement 2. The attempt at a solution I've tried using the relation Cp = T(dS/dT), isolating "T" for T = Cv2(dT/dS) and using the maxwell relations to reduce the derivatives, reaching, T = Cv2/D (dV/dS), but i don't think this is the right way to do solve this problem, i couldn't...

Homework Statement Using double integrals, calculate the volume of the solid bound by the ellipsoid: x²/a² + y²/b² + z²/c² = 1 2. Relevant data must be done using double integrals The Attempt at a Solution i simply can't find a way to solve this by double integrals, i did with triple...

you could try to separate the two halves, upper and lower,and multiply by 2 at the end, let's say we take the upper half of the cylinder, then, z goes from 0 to R*Cos∂ (the angule from Z to R), if you cand find ρ, then 0<θ<2Pi and 0<∂<Pi/2 and you have the volume for the upper part of the...

have you tried to find the volume of the sphere and the cylinder and then subtracting? making two integrals instead of one, might be easier to see, also use wolframalpha to plot the graph, might give you some hint as to what the ranges are. also you can use spherical coordinates for one and...

ρ at pi/2 = 0 and ρ at pi/4 = 0.707 using the equation ρ = Cos(pi/2)/(2-2cos²(pi/2)) = 0 ρ = Cos(pi/4)/(2-2cos²(pi/4)) = 0.707 so it checks, thanks for everything!

actually it was (2-2cos²ϕ) in my notes, damn you keyboard! is it perfect now?

ok, i think i finally understood (thanks to you), the lower limit for ρ is 0, and the upper limit is now ρ = cosϕ/2 + ρcos²ϕ, solving for ρ gives me ρ = cosϕ/(2-cos²ϕ) and the ranges for the spherical coordinates are: 0 < θ < 2Pi pi/4 < ϕ < pi/2 0 < ρ < cosϕ/(2-cos²ϕ) is that it? thanks for...

so now i have 2ρ²cos²ϕ < ρ² < ρcosϕ/2 + ρ²cos²ϕ , and the next step should be isolating ρ to the center of the inequality, right? so i tried dividing for ρ, getting, 2ρcos²ϕ < ρ < cosϕ/2 + ρcos²ϕ but i can't seem to isolate ρ. how should i proceed?

so i got that z/2 < x²+y² < z² becomes z/2 < r² < z², i added z² in the equation and got z/2 + z² < P² < 2z², any thoughts of what should i do now? or isn't this the right way to proceed?

do you have an example of a problem that looks like this one?, i haven't found anything similar in my book or on the internet, so it makes things a little bit difficult to see.

i'll give it another try tomorrow,it's 4am here and I am sleepy(abeit stubborn),and I'm starting to get frustated because I am not getting anywhere and the lack of an actual answer in this book makes things worse. i'll see if things start to get better in the morning. but you are right, that...

sorry, i miss read that, are the ranges Pi/4<ϕ<Pi/2 and 0<ρ<0.7?

i'll take a look at the plot z vs x, and i had already tought about cylindrical coordinates but the question asks it to be in spherical coordinates and not in cylindrical.

i still couldn't find the range for ρ, i think i might be missing something and i don't have the answer to check wheter i am right, could you elaborate a little more on how to find ρ,θ and ϕ? i tried using sqrt(x²+y²) <ρcosϕ< 2*(x²+y²) then, ρsinϕ<ρcosϕ<2*(ρsinϕ)² but i don't know how to...