Recent content by Kreamer

ahh clever

I asked a friend after posting this. They said you use the area of the left side and multiply it by the electric field coming out of the slope as if it was just a rectangle and there was no slope at all. No need for any cosin or angles at all. I tried this and it gave me the right answer but I...

Homework Statement The electric field has been measured to be horizontal and to the right everywhere on the closed box shown in the figure. All over the left side of the box E1 = 90 V/m, and all over the right, slanting, side of the box E2 = 400 V/m. On the top the average field is E3 = 120...

I am sorry my brain is just fried, got to go take the test now. Wish me luck :/

Would the limits still be 0 to 1 for x then -x to 0 for y? I am running on 3 hours of sleep so I feel completely lost sorry. If so switching the order would only add an x to the equation then plug in a 1 for the x and a subtracting the 0 form of the equation giving me a single integral of just...

I am not following. I can't even begin to integrate it, regardless of the limits.

pretty sure the input got messed up and i couldn't fix it. limits are -x to 0 for y and 0 to 1 for x

\int^{1}_{0}\int^{0}_{-x} \frac{ysin(pi*y^2)}{1+y} dydx Not exactly sue how to start this. I know that I need to integrate with respect to y first then use that solution and integrate again with respect to x however I do not believe integrating the initial problem is possible. Is there another...

\int^{1}_{0}\int^{x/2}_{0}\frac{y}{(2y-1)\sqrt{1+y^2}}dydx Most of my attempts at this problem fail pretty quickly. Not even my calculator knows what to do with this one.

"It's so damn hot... milk was a bad choice..." -Ron Burgundy

Ah yes yes I get it now. However I believe it is supposed to be 2, not 4. The equations were Z squared not just Z. But I think I get it now. Thank you! Edit: Also I am not sure why I said set z=0, not what I did, just a bit tired I guess

(x2 + y2)1/2 = z2 (x2 + y2)1/2 = 8 − z2 I substituted then set z=0 and solved for x and y to get x = +/- (y2-16)1/2 and y = +/- (x2-16)1/2 Are these my limits or did I go wrong somewhere along the line?

I am starting to get it more. I am attempting the problem but am having trouble figuring out how exactly you get the right limits of integration.

Just working on some practice problems. I missed a couple classes due to sickness and just need some extra help. If you could walk me through how to do these types of problems that would be amazing. Homework Statement Evaluate the volume of the solid bounded by the surfaces (x2 + y2)1/2 =...

Ahhh I am forgetting the basics! Forgot about rest mass energy. Thank You!