Recent content by City88

Homework Statement Let x be the solution to dx/dt=(pi-x)cos(5x) x(0)=4pi/3 Find the lim x(t) as t goes to infinity. Homework Equations The Attempt at a Solution cos(5x)=0 when 5x=n(pi), so when x=n(pi)/5 which = 4(pi)/3 since we have that initial condition. so, n=20/3...

Homework Statement I have to find the volume of the region bounded by a cone and cylinder. (double integral style!) I usually start off with a sketch, but I can't seem to figure out what this one quadratic surface looks like... It's a cylinder (in 3-dimensional plane): x2+y2=2ay The...

Hi everyone... I was wondering if anyone had subscribed to calc101.com? Most of the site is free, but for step-by-step integration, you need to buy a password. But is it worth it? I'm taking an ad-cal course at the moment, and I would like to be able to double check various integration. Thanks!

Homework Statement Find the equation of the plane through the line of intersection of the planes x-z=1 and y+2z=3 AND perpendicular to the plane x+y-2z=1 Homework Equations The normal of the first plane is n1=(1,0,-1), second is n2=(0,1,2) The Attempt at a Solution I started off by...

Thats what I first thought also...but then I realized the equation was just waaaay too messy for my liking. So then I figured polar coordinates may have to be used, since an ellipse sort of resembled a circle. I just tried doing it again, but by keeping it in terms of x and y...and well it's...

Tiny-tim, my original question is https://www.physicsforums.com/showthread.php?t=243360 I'm trying to solve the double integration, and can't figure out any other way to do it, well other than polar coordinates. thanks for the θ and π!

Even after doing that, I'm still stuck.. x-2 = u = rcos(theta) y-4 = v = rsin(theta) Wouldn't I still have to add (more like multiply) something to the rcos(t)? The radius still isn't circular because the u2 and v2's are divided by 42 and 62.

Homework Statement I'm trying to solve a double integral of a function which is bounded by the ellipse: \frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}} = 1 And I can't figure out how to write this in polar coordinate form, and also what my bounds for theta and radius would be. Homework Equations I...

Actually I just found out that I might have to use polar coordinates. But since the region is not a circle, but an ellipse, how would I be able to write my x and y in polar coordinate form? Normally we have x=rcos(theta), and y=rsin(theta) for a circle. would it be... x=4rcos(theta) + 2...

Ooohh yes, you're right. So does this mean that I have to find the bounds of my integral as functions? Would I have to find the equation of the ellipse in terms of y, and make that my bounds for the y-integral. and then keep the x integral as numbers? like D = {(x,y)| a \leq x \leq b...

Homework Statement I'm not sure how to go about finding the following moments: M_{x}= \int \int\ y dx dy M_{y}= \int \int\ x dx dy Where the region is bounded by the ellipse: \frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}} = 1 Homework Equations Listed above... The Attempt...

Hi there... I'm not sure how to go about finding the following moments: M_{x}= \int \int\ y dx dy M_{y}= \int \int\ x dx dy Where the region is bounded by the ellipse: \frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}} = 1 I tried this several ways. I drew the ellipse and found the bounds to be -2...

Hi, I'm working on a cal III problem involving implicit differentiation. I have to find the second order partial derivative of an implicit function, basically: \partial2f \partialx2 now, I know that for a single order \partialf/\partialx, I would simply use the chain rule property: \partialf =...