Triple Integrals in Solid Tetrahedrons - Solving for Z

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Q
1)double integral over[ (0 to a/sq.root 2),(y to sq.root of (a2-y2)] fn->log(x2+y2)dxdy &a>0

2)int (0to pi/2) d(theta) int(0 to a sin theta) dr int(0 to (a2-r2)/a) r dz

3)evaluate triple integral over V funtion=> z dx dy dz, where V is the solid tetrahedron bounded by the 4 planes..x=0,y=0,z=0&x+y+z=1
 
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please teach me how to draw a parabola if its eqn is given as:
...y2=2x+6
...x2=y+1

if u can't tell me in detail...please just try to give it just in 2-3 lines...Once i knew well to draw all these...but now i can't recollect those...
 


5)what about a question given to find the volume of a hyperboloid?
please try to illustrate it with examples...

6)semicircle over an initial line's theta varies from ____ to _____?

7) what about a paraboloid??

8)triple integral over B fn=>xy(z2) dx dy dz where B is the rectangular box given B={9x,y,z)/0<=x<=1, -1<+y<=2, 0<+z<+3}
 


9) evaluate teriple integral over V, fn=> xy dx dy dz, where V is the solid tetrahedron with vertices(0,0,0),(1,0,0),(0,2,0) and (0,0,3)
10) evaluate triple integral over V fn>x dx dy dz, where V is the paraboloid x=4(y^2)+4(z^2) and the plane x=4...

Can I please have the figures too?...Kindly do it if u can


Try to reply, today or tomorrow(as fast as possible)...am just in need...

Please try to give a general description on each type of question, so that i can do other such types of question easily by referring that...


Thanking you in advance,
ag
 


We're not an answering machine. Start with question 1, show us what you've done so far and where you get stuck, so we can nudge you into the right direction.
 


sorry but thanks for letting me know...I was in a hurry and i...am.
thts why I didnt try explain my soln...
from my next doubt onwards i'll b trying to follow the writing rules of this forum...

Thanks Again
ag:smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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