Respuesta :
Answer:
Z = 8 + 2x2 + 2y2
Convert to polar coordinates
Z = 8 + 2r2
Now theta will go from 0 to pi/2 because it's in the first quadrant.
R will go from 0 to the radius of the circle formed at the intersection of the plane and the paraboloid.
14 = 8 + 2r2
r = sqrt(3)
So r goes from 0 to sqrt(3).
You integrate 14-z where 0<r<sqrt(3) and 0<theta<pi/2.
It is 14-z and not z because just z would give the volume under the paraboloid.
Step-by-step explanation: please go answer my recent question
To answer this question it is necessary:
- Determine the integration limits
- Solve the resulting double integral after changing to polar coordinates.
The solution is:
V = (9/4)× π cubic units
z = 9 + 2×x² + 2×y² and the plane z = 15
To find integration limits we look the interseption poitns so
15 = 9 + 2×x² + 2×y²
6 = 2×x² + 2×y² ⇒ 3 = x² + y² = r² ⇒ r = √3
Then r ⇒ 0 ≤ r ≤ √3
The question is about the volume in the first octant, then
0 ≤ θ ≤ π/2
Now :
x = r×cos θ y = r×sin θ and dA = rdrdθ
V = ∫∫ ( 3 - r²×cos²θ + r²×sin²θ ) ×rdrdθ
V = ∫∫ ( 3 - r² )×r dr dθ
V = ∫ (3 - r² )× rdr ∫ dθ
V = (3/2)×r² - r⁴/4 | ( 0 √3 ) = (9/2) - (9/4)
V = (9/2) × π/2
V = (9/4)× π
V = (9/4)× π cubic units
Related Link :https://brainly.com/question/25172004
