Question

Hey I'm having trouble getting my head around this one could anyone explain how I could evaluate this please :)

Let D={(x,y,z) such that z^2=1+x^2+y^2 and 1 14 years ago

Answer 1

this is how i would do it
\[1<z<3\]
\[1<z^{2}<9\]
\[1<1+x^{2}+y^{2}<9\]
\[0 <y<\sqrt{8-x^{2}}\]
\[\rightarrow -2\sqrt{2} <x<2\sqrt{2}\]
\[\large \int\limits_{-2\sqrt{2}}^{2\sqrt{2}}\int\limits_{0}^{\sqrt{8-x^{2}}}\sqrt{1+x^{2}+y^{2}}dydx\]

Answer 2

Thanks very much for the reply, but in your line 4 shouldn't there be an -x^2 on the left of the inequality and if so how would you solve? (I'm not trying to be picky This stuffs driving me up the walls so your help is very much appreciated)

Answer 3

no i agree with you, but you cant have a neg inside radical
so you have to adjust the min value of y
y must be greater than 0 to satisfy 0<x^2 +y^2 < 8

Answer 4

surely y could be less than zero because you are squaring it?

Answer 5

ok yeah that won't work, well to satisfy the equation --> 0<x^2 +y^2 < 8
both x and y are bounded by -2sqrt2 to 2sqrt2
maybe this then:
\[\int\limits_{-2\sqrt{2}}^{2\sqrt{2}}\int\limits_{-2\sqrt{2}}^{2\sqrt{2}} \sqrt{1+x^{2}+y^{2}}dydx\]

Answer 6

What about a change of variables, not sure how we'd manage it but it might make the problem a bit easier? maybe