To find the volume inside the given region, we need to first find the intersection between the two planes.
From the equation x + y = z, we can rearrange it to z - x - y = 0.
Substituting this equation into x^2 + y^2 = 9, we get (z - x - y)^2 + y^2 = 9.
Expanding the equation, we have z^2 + x^2 + y^2 - 2xz - 2yz + 2xy = 9.
Since we only need to consider the region above the plane z = 0, we can simplify the equation to x^2 + y^2 + 2xy = 9.
We can rewrite this equation as (x + y)^2 = 9, which means x + y = ±3.
To find the intersection points, we can solve the following system of equations:
x + y = 3
x + y = -3
Solving for x and y in each equation, we get:
x = 3 - y
x = -3 - y
Setting these two expressions for x equal to each other:
3 - y = -3 - y
Solving for y:
2y = 6
y = 3
Substituting this value of y back into one of the equations, we get:
x = -3 - y
x = -3 - 3
x = -6
Therefore, the intersection point is (-6, 3).
To find the volume inside the region, we need to calculate the double integral:
∬(x^2 + y^2) dA
Over the region R, which is the circle with radius 3 centered at the origin.
Since the region is symmetric, we can just consider the area of the upper half of the circle.
Converting to polar coordinates, we have:
x = r cosθ
y = r sinθ
The equation x^2 + y^2 = 9 becomes:
(r cosθ)^2 + (r sinθ)^2 = 9
r^2(cos^2θ + sin^2θ) = 9
r^2 = 9
So, the limits of integration for r are 0 and 3.
The limits of integration for θ are 0 and π.
The double integral becomes:
∬(r^2) r dr dθ
Integrating with respect to r first, we have:
∫[0, π] ∫[0, 3] (r^3 / 3) dr dθ
Integrating with respect to θ, we get:
(1/3) ∫[0, π] [(3^4 / 4)] dθ
Simplifying, we have:
(81/12) (∫[0, π] dθ)
(81/12) (π - 0)
(81π / 12)
Therefore, the volume inside the given region is (81π / 12) cubic units.
Find the volume inside the calendar x^2+y^2=9 above the plane z=0 and below plane x+ y=z
1 answer
1 answer