# 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=∫∫_{D}ydS

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.