It is a paraboloid that increase where x increase , at x=0 ( Vertex) , so we need a limit for x .-
Since f(x,y,z) = x-y^2- z^2 =0 a normal to the surface is Nabla F
N= 1 i-2y j-2zk and a unit normal is= N/INI , so dS projected over YZ plane is
I dot i I dS = dA
I dot iI = 1/ INI
dS= INI dA
dS= sqrt (1+4y^2+4z^2) dA
Taking cylindrical coordinates
y=rcosT
z= rsinT
x=x
dA= rdrdT
dS= sqrt ( 1+4r^2 ) dA
INT_S yz dS = INT_R r^2sinTcosT sqrt (1+4r^2) rdrdT
= INT r^3sinTcosT sqrt (1+4r^2)drdT
If u = sinT , du =cos dT dT
=(1/2) sin^2T INT r^3 sqrt (1+4r^2)dr
0
so,
INT _S yz dS = (1/2) INT r^3 sqrt (1+4r^2)dr
We need a limit for x , x=xo , so the region will be a circle with radius R= sqrtxo and
0
Since f(x,y,z) = x-y^2- z^2 =0 a normal to the surface is Nabla F
N= 1 i-2y j-2zk and a unit normal is
I
I
dS= INI dA
dS= sqrt (1+4y^2+4z^2) dA
Taking cylindrical coordinates
y=rcosT
z= rsinT
x=x
dA= rdrdT
dS= sqrt ( 1+4r^2 ) dA
INT_S yz dS = INT_R r^2sinTcosT sqrt (1+4r^2) rdrdT
= INT r^3sinTcosT sqrt (1+4r^2)drdT
If u = sinT , du =cos dT dT
=(1/2) sin^2T INT r^3 sqrt (1+4r^2)dr
0
INT _S yz dS = (1/2) INT r^3 sqrt (1+4r^2)dr
We need a limit for x , x=xo , so the region will be a circle with radius R= sqrtxo and
0
1
keywords: yz,bounded,dS,Let,first,octant,integral,double,surface,in,Calculate,be,by,the,Let S be the surface x=y^2+z^2, in the first octant. Calculate double integral (yz*dS) bounded by S.