Recent content by radji

Thank you for your observation. I was able to correct the nodes and solve for each unknown current. Even found out how to solve using only one node!

Thank you for your reply. This is a lab experiment so I have to follow (i.e. use) the circuit diagram I am given. You are absolutely right, I need only the two loops and one junction at B to solve for the 3 currents. I am more looking to see if my loop equations are setup correctly. Then I...

Homework Statement I am attempting to solve the circuit for three currents (I1, I2, I3). I am starting with I1 in terms of I2, then using a third equation to solve for I2 in terms of E and R. I wrote my loops at the bottom of the page. I am using Loop 1 @ B, Loop 2 @ B, and Junction at B to...

Thank you for your help. I ended up using a website to decompose it to partial fractions. It was a very messy solution. ##u=\frac 1 {288} (61\sqrt 2 - 15 sinh^{-1}(1))##

You're absolutely right. That leaves me with tan6u * sec3u. Wondering how that can be integrated...

Thank you for this. I did come up with a reasonable answer using 1+tan2u=sec2u. It still was not the correct answer, but at least I know it how to integrate such a function now. I've attached my work so you all can let me know where else I am miscalculating. Thank you!

2 cos2(u) + 1. I've substituted x for 2 cos2(u) + 1. It gets rid of the (x+1)1/2, but I also get that identity within the 2/3x5/2. And I can't get that expression to simplify out now.

Yes sir, I have. But I couldn't get the √(1+x) to change to a single trig function since its 1+x.

Homework Statement It is evaluating a surface integral. Homework Equations ∫s∫ f(x,y,z) dS = ∫R∫ f[x,y,g(x,y)]√(1+[gx(x,y)]2+[gy(x,y)]2) dA The Attempt at a Solution I set z=g(x) and found my partial derivatives to be gx=√x, and gy=0. I then inserted them back into the radical and came up...