Recent content by quickclick330

okay thanks...I'll ask the teacher then, its an online submission homework so maybe somethings wrong. hopefully.

okay so completing the square gives me y = x^2 +8x + 20 + 16 - 16 y = (x^2 + 8x +16) + 4 then factor... y = (x + 4)^2 + 4 which is exactly what I got before?

Change the equation to standard form. y = x^2 + 8x + 20 I thought this was the standard form for parabolas?? I tried this as the answer but it said it was wrong y = (x+4)^2 +4 Thanks for the help! :-)

[SOLVED] More trig substitution help... I've looked at this problem about 3 times and still can't figure it out...where identity did they use to substitute out the part in the red box? Thanks for the help

Hi I'm having trouble understanding how they simplified this integration using trig substitutions...I really don't know what identities they used to make these substitutions or the strategy behind why this substitution was made(particularly the 3 steps in the red box). Obviously it works but I...

never mind, i solved it finally

yes, sorry, it copy-pasted like that

Consider the given curves to do the following. x = 3 + (y-2)**2, x = 4 Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. V = ?? **************** I set up the problem like this... V...

Consider the solid S described below. The base of S is a circular disk with radius 3r. Parallel cross-sections perpendicular to the base are squares. Find the volume V of this solid. I tried this... A(x) = pi*r^2 = pi*(3r)^2 = pi*9r^2 V(x) = integral from 0 to 3r(pi*9r^2 dx) =...

Harmonic Oscillator 9. An ideal harmonic oscillator has energy levels separated by a constant 0.05 eV. That is, the difference between the ground state and the first excited state of this oscillator is an energy of 0.05 eV. This single oscillator is brought into contact with a large solid...

okay, thanks

I do have one more question...how do u recalculate the new moment of inertia?

Thanks!

so initial and final angular momentum should equal since there is no external torque on the system right?

The moment of inertia?