Respuesta :
Answer:
the integration limits are:
0≥θ≥π/2
1/cos φ≤ρ≤√10
0≥θ≥ cos⁻¹ (1/√10)
Step-by-step explanation:
the triple integral would be initially
[tex]V = \int\limits^{z2}_{z1}\,\int\limits^{y2}_{y1} \int\limits^{x2}_{x1}\ dxdydz[/tex]
for
x²+y²+z²≤10
that should be converted into spherical coordinates in order to solve the integral more easily. Then:
x= ρ* cosθ * sinφ
y= ρ* senθ * sinφ
z= ρ* cosφ
then
x²+y²+z²= ρ²
since ρ= distance from the origin , θ = angle with respect to the x axis in the x-y plane , φ= angle with respect to the z axis . The integration limits are
0≥θ≥π/2 (positive x and y values, first octant )
since
x²+y²+z²= ρ² and x²+y²+z²≤10 , then
ρ≤√10
also
x²+y²+z²= ρ , cut by z=1 , then
z= ρ * cos φ = 1 → ρ=1/cos φ
therefore
1/cos φ≤ρ≤√10
since ρ≤√10 also √10=ρ=1/cos φ and φ= cos⁻¹ (1/√10) , thus
0≥θ≥ cos⁻¹ (1/√10) ( the positive x and y values, first octant)
thus the integration limits are
0≥θ≥π/2
1/cos φ≤ρ≤√10
0≥θ≥ cos⁻¹ (1/√10)
then considering the Jacobian for the transformation to spherical coordinates
J= ρ²*sin φ
the integral will be
[tex]V = \int\limits^{{cos}^{-1} (1/\sqrt{10})}_{0}\,\int\limits^{\sqrt{10}}_{1/cos(phi)} \int\limits^{2\pi }_{0}\ (rho)^{2} *sin (phi) d(theta)d(rho)d(phi)[/tex]
