The domain bounded by the surface of a paraboloid z=2-x^2-y^2 and that of a cone z^2=x^2+y^2 is given by D={x,y,z:x^2+y^2

Question

The domain bounded by the surface of a paraboloid z=2-x^2-y^2 and that of a cone z^2=x^2+y^2 is given by D={x,y,z:x^2+y^2<=1, sqrt(x^2+y^2)<=z<=2-x^2-y^2} can someone explain how to get the upper limits for x,y,z please

turingtest · Answer

change this to cylindrical coordinates

anonymous · Answer

I need to do this triple intergral intergration and just need to understand how to get the upper limits for x,y,z in order to intergrate

anonymous · Answer

can you explain this please if you are able to?

turingtest · Answer

in rectangular coordinates your bounds are

sqrt(x^2+y^2)<=z<=2-x^2-y^2
0<=y<=sqrt(1-x^2)
0<=x<=1

but that is an ugly integral, I strongly recommend switching to cylindrical coordinates

anonymous · Answer

ok, thanks for the the bounds but I would just appreciate it if you could explain how you came up with the bounds, sorry to be so thick.

turingtest · Answer

sorry I meant

sqrt(x^2+y^2)<=z<=2-x^2-y^2
-sqrt(1-x^2)<=y<=sqrt(1-x^2)
0<=x<=1

you get the bounds for z directly from the area:
sqrt(x^2+y^2)<=z<=2-x^2-y^2

you are also given that it is over the region
x^2+y^2=1

solve this for one variable:
-sqrt(1-x^2)<=y<=sqrt(1-x^2)

since the argument under the square root cannot be negative we see that x cannot be greater than 1,hence the bounds on x are

0<=x<=1

anonymous · Answer

thank you sooo much for that explanation

anonymous · Answer

now how would you change it to polar coordinates?

turingtest · Answer

cylindrical coordinates:
$$x=r\cos\theta$$$$y=r\sin\theta$$$$z=z$$$$dA=rdzdrd\theta$$use the symmetry of the object to find the bounds on r and theta

anonymous · Answer

ok

anonymous · Answer

thank you!

turingtest · Answer

welcome!