- #1
mugzieee
- 77
- 0
i have to setup a doble integral to find the volume of the solid bounded by the graphs of the equation.
x^2+z^2=1, and y^2+z^2=1
z=sqrt(1-x^2)
z=sqrt(1-y^2)
then substituting in z=sqrt(1-y^2) into x^2+z^2=1, i got y=x.
so when i setup a double integral
for the dy i get integral of(sqrt(1-y^2)) and the limits are from 0-x
and for the limits of he dx i get 0-1.
the way i got the lim its of x was by solving the equation x^2+z^2=1 for x by making z=0, but I am not sure if that's right.
and finally when i integrate my double integral, I am getting a wrong answer.
x^2+z^2=1, and y^2+z^2=1
z=sqrt(1-x^2)
z=sqrt(1-y^2)
then substituting in z=sqrt(1-y^2) into x^2+z^2=1, i got y=x.
so when i setup a double integral
for the dy i get integral of(sqrt(1-y^2)) and the limits are from 0-x
and for the limits of he dx i get 0-1.
the way i got the lim its of x was by solving the equation x^2+z^2=1 for x by making z=0, but I am not sure if that's right.
and finally when i integrate my double integral, I am getting a wrong answer.