volume of solid bounded by cone of x^2+y^2=z^2 and the paraboloid of x^2+y^2=z ? and someone draw the graph and explain thanks....use triple integral

Question

volume of solid bounded by cone of x^2+y^2=z^2 and the paraboloid of x^2+y^2=z ?  and someone draw the graph and explain thanks....use triple integral

anonymous · Answer

http://www.wolframalpha.com/input/?i=plot+z%3Dsqrt%28x^2%2By^2%29+and+z%3Dx^2%2By^2

anonymous · Answer

wow i can now visualize it clearly @cinar but now for the triple integral formula i dun know the region of integration lies could u explain to me and is it cone minus paraboloid or vise versa? ty....

anonymous · Answer

examine example 5 http://www.math24.net/calculation-of-volumes-using-triple-integrals.html

anonymous · Answer

wow kk steps are explained clearly but do u know which minus which?? cone minus paraboloid or paraboloid minus cone that confusing me only .....

anonymous · Answer

choose one of them we know that volume can not be negative, if it is negative then it is versa..

I guess second one has larger volume, you can see it at graph

anonymous · Answer

I forgot these stuff sorry..

turingtest · Answer

x^2+y^2=z^2 means that z=sqrt(x^2+y^2) 
x^2+y^2=z means that     z=x^2+y^2
so z will have a greater value in the cone than the parabaloid, hence it is on top

anonymous · Answer

you are saying opposite of what you thinking

turingtest · Answer

you are right, I meant it has a greater value in the parabaloid, sorry

anonymous · Answer

hmm means its paraboloid minus cone? @TuringTest

turingtest · Answer

yes, because$$\sqrt{x^2+y^2}\le x^2+y^2$$anyway, this should be easier in cylindrical coordinates $$x^2+y^2=r^2$$
I have to go take a Spanish test, good luck, I'll check on this when I return :D

anonymous · Answer

if i do both paraboloid minus cone and vise versa will i get exact value but diff sign one positive one negative? @TuringTest

anonymous · Answer

http://www.math.ubc.ca/~malabika/teaching/ubc/fall08/math263/hw6-solution.pdf

anonymous · Answer

http://www.wolframalpha.com/input/?i=%28x^2%2By^2%29^2-x^2-y^2%3D0

anonymous · Answer

http://www.physicsforums.com/showthread.php?t=15266

turingtest · Answer

yes I think you could change the sign at the end
we need the intersection of these two surfaces
converting this to cylindrical coordinates$$x^2+y^2=r^2=z^2\to z=\pm r$$$$x^2+y^2=r^2=z$$setting the z of both equations equal we get$$r=r^2\to r=0,1$$hence our bounds are$$r\le z\le r^2$$$$0\le r\le1$$$$0\le\theta\le2\pi$$and you can now integrate