Question

can someone help me with this question please !
-Compute the volume of gama bounded by : z=x^2+y^2, 4-z=x^2+y^2

Answer 1

\[\gamma:=\left\{(x,y,z)~:~x^2+y^2\le4,~x^2+y^2\le z\le4-x^2-y^2\right\}\]
So as an integral,
\[V=\int\int\int_\gamma dV=\int_{-2}^2\int_{-2}^2\int_{x^2+y^2}^{4-x^2-y^2}dz~dy~dx\]

Answer 2

then i just compute the integral from z then y then x and that's it ? right ?

Answer 3

Yes, leftmost two integrals are with respect to x and y, rightmost integral is with respect to z.

Answer 4

z = x^2 + y^2 is a paraboloid (notice x^2+y^2 = r^2 is a circle)
it looks like this

Answer 5

4-z=x^2+y^2
can be written as
z = -(x^2+y^2) +4

It is upside down relative to the other paraboloid, and shifted up by 4