 # Ann_M

Senior Members

32

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Quark

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1. Yes, I see it. ok, meaning my comps playing up
2. Solve this question. what question, there is no question showing up, can any one else see it?
3. It's not fixed.....I'm The Grinch!!! Actually it's Ginch, but close enough!!! lol well im Gnn_m hmmmmm well both the names sound good to me!!!
4. yeah, so wheres the evidence MR Fluent in lies?
5. ## Maths Problem 3: Peaks and Troughs

lol keen
6. nice to know about his other languages as well!!! i prefer visual basic although at the moment ive just touched upon it. Done a bit of SQL, that was not bad.
7. okey dokey i have worked out the limits according to me i get: 1st integral: 0 : 1 2nd integral: 0 : 2-2x 3rd integral: 0 : (2-y-2x)/2 i have worked this out by dz dy dx and i get the volume to be 1. can some one verify that 4 me. thanx oh by the way Neurocomp2003 i am really appreciating ur help!!! thanx alot
8. yep i get the diagram, to find the equation of the plane, is it possible to let the two equations equal each other, ie, 1-x=1-(y/2) simplified to z = 2-y-2x as this connects x and y.
9. i hope u dont mind, and this is ok with u.
10. i think i should explain my situation, i have come back to maths after some time, and im in a situation where i cannot ask any one as this is a course is studied from home. no one i know, knows integration, therefore i was getting help from u guys, as i then know if im heading the right way.
11. i need to find the volume bounded by x + z = 1 and y + 2z=2 in the positive quadrant ive worked out the co-ordinates which are: x(1,0,0) y(0,2,0) Z(0,0,1) how do i find out the limits to put for the three integrals. i think the first integral is going to be: 0 to (1-z) 2nd integral: 0 to (2-2z)/2 but im not sure about these two either can some clarify for me if im going about this the right way. im not sure about the third one. can any one help. thanx
12. i dont understand, i get (-2R^2)^3/2 - (R^2)^3/2 after i substitued 0:R and then wen i do the final integral i get: 1/3 int 0 to 2pi (-2R^2)^3/2 - (R^2)^3/2 d theta after this im confused, how do u go about subtracting that.
13. yep i get that and i also get -2/3 outside the integrals, which i multilply right at the end. but even with that u dont get R^3 in the final answer. this is wat i get wen i have to put the limits of 0 and R in: 1/3 int(0 to 2 pi {(R^2 - r^2)^3/2)} and the limits 0 and R is dis wat ne of u get wen u tried it.
14. help me den please im sooo confused i get 8/3 pi r ^2 i dont know wat im doing wrong, try it ur self and c wat u get.
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