Evaluate the surface integral.
double integral ydS
S is the part of the paraboloid
y = x2 + z2
that lies inside the cylinder
x2 + z2 = 1
This is a problem where the paraboloid projects a circle of radius 1 (represented by the region D) on the x-z plane.
y=0 when x=z=0
y=1 when x²+z²=1
So let
I=∫∫DydS
It may be changed to cylindrical coordinates
where dS=rdrdθ, and using
y=x²+z²=r²,
I=∫[0,2π]∫[0,1]r²*rdrdθ
=2π[r^4/4][0,1] after separating r & θ
=2πr^4/4
=πr^4/2
what is r ? the answer doesn't have r in it
x^2 + z^2 = r^2 is a circle with radius r
read what mathmate wrote for you.
To evaluate the surface integral, we need to find the parametric equation of the surface S and then use a double integral to calculate the flux across the surface.
The surface S is defined by the equation of the paraboloid y = x^2 + z^2, and it lies inside the cylinder x^2 + z^2 = 1.
To find the parametric equation of the surface, we can use cylindrical coordinates. Let's parametrize the surface as:
x = r*cos(theta)
y = r^2
z = r*sin(theta)
Here, r is the radial distance and theta is the angle in the xy-plane.
Using the equation of the paraboloid, we can express r^2 in terms of y:
r^2 = y
Substituting x, y, and z into the equation of the cylinder, we have:
(r*cos(theta))^2 + (r*sin(theta))^2 = 1
r^2*cos^2(theta) + r^2*sin^2(theta) = 1
r^2 = 1
Since r^2 = y, we have:
y = 1
Therefore, the parametric equation of the surface S is:
x = r*cos(theta)
y = 1
z = r*sin(theta)
To evaluate the surface integral, we convert it into a double integral over the region in the xy-plane that corresponds to the projection of the surface S. In this case, the region is a circle with radius 1.
Now, we can set up the double integral:
∬(S) y dS = ∬(R) (1) sqrt((dx/dtheta)^2 + (dy/dtheta)^2 + (dz/dtheta)^2) dA
Here, R represents the projection of the surface S onto the xy-plane, and dA is the differential area element in the xy-plane.
The partial derivatives dx/dtheta, dy/dtheta, dz/dtheta are:
dx/dtheta = -r*sin(theta)
dy/dtheta = 0
dz/dtheta = r*cos(theta)
Therefore, the square root term becomes:
sqrt((dx/dtheta)^2 + (dy/dtheta)^2 + (dz/dtheta)^2) = sqrt((-r*sin(theta))^2 + (r*cos(theta))^2) = sqrt(r^2) = r
The double integral becomes:
∬(S) y dS = ∬(R) (1) r dA
Since y = 1 and the integral is over a constant function, the value of the surface integral is equal to the area of the region R:
∬(S) y dS = ∬(R) r dA = A(R)
Where A(R) represents the area of the region R.
In this case, the region R is a circle with radius 1, so its area can be calculated using the formula:
A(R) = π * (radius)^2 = π * 1^2 = π
Therefore, the value of the surface integral is π.