We want to evaluate the given
surface integral, where the surface S
is a helicoid with the vector equation given
by R of u comma v, and u is on the closed interval
from zero to four, and v is on the closed interval
from zero to two pi.
Were surface S given parametrically,
or by a vector-valued function R of u comma v as shown here,
that is defined over a region R in the uv-plane,
so the region R will be the projection of R of u comma v
onto the uv-plane, the surface integral
of f of x comma y comma z over the surface S
written as the double integral over the surface S
of f of x comma y comma z, integrated with respect to S
is equal to the double integral over the region R
of f of u comma v times
the magnitude of the partial derivative
of r with respect to u
crossed with the partial derivative of r with respect to v
times differential A.
Just notice how to evaluate with the surface integral,
we write f of x comma y comma z as a function of u and v
using r of u comma v, and differential s is equal to
the magnitude of this cross-product, differential A.
So going back to our problem, let's first take a look
at the surface S in space.
So we're integrating the inter gram function
over this surface, and notice how if we project this
onto the uv-plane or look down on the uv-plane,
we can see that region R in the uv-plane,
is a circle centered at the origin
with the radius of four.
So over here we have a graph of the region R
and the uv-plane which we'll use later
to determine the limits of integration.
Let's first write the inter gram function
as a function of u and v.
We're looking at r of u comma v,
x equals u cosine v,
y equals u sine v, and z equals v.
So f of u comma v would be the square root
of the quantity one plus x-squared,
would be the square of u cosine v
plus y-squared would be the square of u sine v.
Let's go ahead and simplify.
We'd have one plus u-squared cosine-squared v
plus u-squared sine-squared v.
We notice how here if we factor out u-squared,
we'd have f of u comma v equals the square root
of the quantity one plus u-squared
times cosine-squared v plus sine-squared v,
so f of u comma v is just the square root
of the quantity one plus u-squared.
Now for our next step, let's go ahead and find
the partial derivatives of r with respect to u and v.
Let's do this on the next slide.
So the partial derivative of r with respect to u
would have an x component that's a partial derivative
of u cosine v with respect to u which would be cosine v.
The y component is a partial of u sine v with respect to u
which would be sine v, and the z component
is the partial of v with respect to u which would be zero.
Then we're going to find the partial with respect to v.
So the x component is going to be the partial of
u cosine v with respect to v
which would be u times negative sine v,
or negative u sine v.
The y component is the partial of u sine v
with respect to v which would be u cosine v,
and the z component is the partial of v with respect to v
which would be one.
So then we want to find the cross-product
of these two partial derivatives.
We'll determine this cross-product
using a three-by-three determinant as shown here,
where the first row of the unit vector is i, j, and k,
the second row are the components
of the partial with respect to u,
the through-row components of the partial with respect to v.
Using expansion by minors
for this first two-by-two determinant,
we eliminate the row and column of the i vector,
so we'd eliminate row one, column one,
leaving us these four elements for the first determinant.
For the second determinant, remember we have a minus here,
and then we'd eliminate the row and the column
of the j-vector, so we eliminate row one column two,
leaving us with these four elements.
And then finally for the last two-by-two determinant,
we eliminate the row and column of the k-vector,
to eliminate row one, column three,
leaving us with these four elements.
Now we evaluate each two-by-two determinant
by determining this product minus this product.
So for the x-component, we have sine v times one
minus zero, so we have sine v times i,
and then we have minus cosine v times one minus zero.
So we have minus cosine v j plus,
for the z-component we have u cosine-squared v
minus negative u sine-squared v,
which becomes u cosine squared v
plus u sine squared v times k.
But again, notice how if we factor out the u here,
we have u times the quantity cosine-squared v,
plus sine-squared v, which simplifies
into just u times one, which is just u.
So in component form, the cross-product
is angle bracket sine v comma negative cosine v
comma u, angle bracket.
And now we want to find the magnitude of this cross-product
which I've also already set up.
The magnitude of the cross-product is equal to
the square root of the square of sine v
plus the square of negative cosine v plus u-squared.
We should recognize this again,
cosine-squared v plus sine-squared v
is equal to one, and therefore, the magnitude
of the cross-product is equal to
the square root of the quantity one plus u-squared.
So going back to our first slide again,
the magnitude of the cross-product
of the partial derivatives
equals the square root of the quantity one plus u-squared.
So now we can set up the double integral shown here.
With the given surface integral
is equal to the double integral
over the region R which we'll come back to,
of f of u comma v which is the square root
of the quantity one plus u-squared
times the magnitude of the cross-product
which is also the square root
of the quantity one plus u-squared.
And then for differential A,
let's use the order of integration du dv.
So using the region here, or just the intervals given
for u and v, the limits of integration for u
are going to be from zero to four
because we've already set the circle as four.
Notice how it's also given here.
I'm going to trace out this region of the uv-plane,
v would have to be from zero to two pi
all the way around the circle which is also given here.
Let's go ahead and evaluate this on the next slide.
Notice how this product simplifies nicely
to one plus u-squared, so this gives us
a double integral of the quantity
one plus u-squared du dv.
And now we integrate with respect to u.
So we'd have u plus u to the third divided by three,
or 1/3 u to the third.
We now need to find big F of v minus big F of a.
So when u is four we have four plus 1/3 times four-cubed,
minus when u is zero we have zero plus 1/3
times zero to the third.
So here we're going to have four plus 64/3 minus zero.
Four plus 64/3 is equal to 76/3.
And now we need to integrate with respect to v,
so we have 76/3 v.
Performing substitution we have 76/3 times
the quantity two pi minus zero.
So we have 76/3 times two pi
which is equal to 152/3 pi,
which would be the exact value of the surface integral,
or the decimal approximation,
this would be approximately 159.1740.
I hope you found this helpful.