Recent content by Pindrought

Thank you very much!

Reading a book about 3d math, and I am confused as to what happened on this Vector Cross Product problem. I'm thinking there was just an error that wasn't caught. For the first row, instead of (3)(8)-(-4)(-5) shouldn't it have been (3)(8)-(4)(-5) and had the same displayed result of 44? And for...

The integral of 1/(1-x) is -(ln(1-x)) which = -ln(1-x) It is negative, because the derivative of the inside (1-x) is -1, so you have to divide by the -1 when finding the integral, which would in this case just make it negative.

I'd love it if someone could verify whether or not I did this problem correctly. A stained-glass window is a disc of radius 2 (graph r=2) with a rose inside (graph of r=2sin(2theta) ). The rose is red glass, and the rest is blue glass. Find the total area of the blue glass. So I set...

I'm really stumped here as usual. Here is what I've managed to figure out. I'm given two equations. r=2 r=4cos(theta)I converted them both to rectangular coordinates to get an idea of what the graph would look like. I need to find either the area in red or the area in green. (In this case...

I can't figure out what I did wrong, because I got ~12.566 for my integral and I got ~4.9 for the rectangular coordinate integral you posted.(Wondering) Do you have any good links to example problems for this type of problem?

I'm also trying to figure this problem out. I tried converting both equations to rectangular form, and finding where they intersect. I got the points of intersection were sqrt(3) and -sqrt(3) I set up my integral as $$\int^{\sqrt{3}}_{-\sqrt{3}} 2*\sqrt{4-x^2} -2 \, dx$$ Am I wrong in...

Re: Work done on a frustrum Thanks for clearing that up. I only have a question about one thing that confuses me. for density, $$\rho$$, should it be just the density of water (62.5lb/ft^3) or should it be the density of water*gravity (62.5lb/ft^3)*32 ? I would think it is just 62.5lb/ft^3...

Re: Work done on a frustrum That's what I thought I did when I said y=2+(1/5)x which I plugged in for the radius. Still can't figure out where I'm messing up.

Hopefully this helps. EDIT: Fixed Image Get it now?

Re: Work done on a frustrum Am I misunderstanding or did you interpret it as my top radius was 2 and my bottom radius was 1? It's supposed to be that my upper radius is 4 and my bottom radius is 2.

Re: Work done on a frustrum $$\int^5_0 \pi (2 + \frac{1}{5} x)^2 * (11 - x) * 2000 \, dx$$ I was thinking i'd set it up like this. Would someone mind just telling me if this is wrong so I can try again if it is?

Okay so please bear with me. I'm dying here. Here's a diagram of the situation I have. I set up the graph to be like so for what I would be rotating to get the volume. (Edit: I meant to put y = 2+1/5x) I'm really stuck here though. I know I need to pump the water to 11ft from it's current...

Thanks a lot Chris L T521, you really helped me understand what I was doing wrong.

Okay so I'm working on this problem $$\int \frac{x^2}{\sqrt{4 - x^2}} \, dx$$ I do a substitution and set $$x={\sqrt{4}}sinu$$ I get to this step fine $$\int 4sin(u)^2$$ I know that u = arcsin(x/2) so I don't see why I can't just substitute in u into sin(u)? I tried this and I got $$\int...