Question

(1 point) Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2+y2+z2≤10 cut off by the plane z=1 and restricted to the first octant. (In your integral, use theta, rho, and phi for theta, ???? and phi, as needed.)

Answer 1

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

$V = \int\limits^{z2}_{z1}\,\int\limits^{y2}_{y1} \int\limits^{x2}_{x1}\ dxdydz$

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

$V = \int\limits^{{cos}^{-1} (1/\sqrt{10})}_{0}\,\int\limits^{\sqrt{10}}_{1/cos(phi)} \int\limits^{2\pi }_{0}\ (rho)^{2} *sin (phi) d(theta)d(rho)d(phi)$