Question

Write a triple integral including limits of integration that gives the volume of the cap of the solid sphere x2+y2+z2≤34 cut off by the plane z=5 and restricted to the first octant. (In your integral, use theta, rho, and phi for θ, rho and ϕ, as needed.) What coordinates are you using? (Enter cartesian, cylindrical, or spherical.) With a= , b= , c= , d= , e= , and f= , Volume = ∫ba∫dc∫fe

Answer 1

Answer:

a = 0

b = $\frac{\pi }{4}$

c = 0

d = 3

e = 5

f = $\sqrt{34-r^2}$

Step-by-step explanation:

Equation of the solid sphere = x^2 + y^2 + z^2 ≤ 34 ------- (1)

at Z = 5

since the bottom of the sphere(z) is flat = 5 we will use cylindrical coordinates

concentrating in the first octant as mentioned in the question

at Z = 5 , equation 1 becomes :

  x^2 + y^2 + 25 ≤ 34 = x^2 + y^2 = 9  

hence the radius around the xy axis = $\sqrt{9}$ = 3

that means the radius is : 0 ≤ r ≤ 4 ,  0 ≤ ∅ ≤ $\frac{\pi }{4}$

next we have to find the upper bound of: Z = ±$\sqrt{34-r^2}$ we will pick out only the positive

5 ≤ z ≤ $\sqrt{34 - r^2}$

therefore for the  Volume = ∫ba∫dc∫fe

a = 0

b = $\frac{\pi }{4}$

c = 0

d = 3

e = 5

f = $\sqrt{34-r^2}$