We want to find the surface area
of the helicoid, or spiral ramp,
with the vector equation shown here,
over the given region.
Let's first take a look at this graphically.
So here's the graph of the vector-valued function.
Our goal here is to find the surface area
of this helicoid, or spiral ramp.
So because the surface is given
as a vector-valued function,
we'll be using this formula here
to find the surface area.
So notice how we'll have to find
the partial derivatives with respect to u and v,
then find the cross product,
then find the magnitude
before we evaluate the double integral.
So let's work on finding our partial derivatives.
To find the partial of r with respect to u,
we differentiate each component with respect to u,
treating v as a constant.
So the derivative of u cosine v with respect to u
would just be cosine v,
the derivative u sine v with respect to u
would be sine v,
and the derivative of v with respect to u
would be zero.
And now for the partial of r with respect to v,
we differentiate with respect to v,
treating u as a constant.
So the derivative of u cosine v
with respect to v would be u times negative sine v,
or negative u sine v,
the derivative of u sine v with respect to v
would be u times cosine v,
and the derivative of v with respect to v
would just be one.
And now we need to find the cross product
of these two vector-valued functions,
so the partial of r with respect to u
crossed with the partial of r with respect to v
is equal to the three by three determinant,
where the first row would be i, j, k.
The second row would be the components of
the partial of r with respect to u,
so we have cosine v, sine v, zero.
The third row is going to be the partial of r
with respect to v, so we have negative u sine v,
u cosine v, and one.
Let's go ahead and evaluate this on the next slide.
So, using the expansion by minors method,
for this first two by two determinant
we eliminate row one and column one,
so we'd have sine v, zero, u cosine v, and one.
For the second two by two determinant
we eliminate row one and column two,
which would give us cosine v, zero,
negative u sine v, and one.
And for the last two by two determinant,
we eliminate row one and column three,
so we have cosine v, sin v, negative u sine v,
and u cosine v.
So the x component is sin(v) time 1 minus
zero u cosine v.
That's going to be sine v.
For the y component the determinant is equal to
cosine v minus zero, that's cosine v,
but because of the minus here we have
negative cosine v.
And for the z component we have
u cosine squared v minus negative u sine squared v,
which is u cosine squared v plus u sine squared v.
Notice how this is going to simplify.
If we factored out u we'd have u times the quantity
cosine squared v plus sine squared v,
and cosine squared v plus sine squared v
is just equal to one,
so the z component simplifies nicely to just u.
So the partial of r with respect to u
crossed with the partial of r with respect to v
is equal to the vector-valued function
with the x component of sine v,
the y component of negative cosine v,
and the z component of just u.
So, going back to our first slide,
now that we have the cross product,
now we need to find the magnitude
of this cross product,
and let's do this on the next slide.
So, we'd have the square root of sine v squared,
plus negative cosine v squared, plus u squared.
And notice how this is going to simplify nicely,
and we'll have the square root of,
this is sine squared v, plus,
here we'll have cosine squared v,
plus u squared, and of course this is equal to one,
so the magnitude of the cross product
is equal to the square root of one plus u squared.
So going back to our first slide again,
we now have all the information we need
in order to set up this double integral
to find the surface area of the helicoid.
The surface area, s, is equal to the double integral
of, again, the magnitude of the cross product
is equal to the square root of one plus u squared.
Now, for differential A let's use du dv.
Now, for the limits of integration for u
we integrate from zero to one,
and for v we integrate from zero to pi.
Now let's go ahead and evaluate this on the next slide.
When we integrate with respect to u, though,
it's very challenging to find this anti-derviative,
so we're going to use the integration form
that's shown here below, from the integration table.
But to make it better fit this formula,
let's rewrite this as the quantity u squared plus one
rather than one plus u squared.
Looking at the integration formula below,
notice how a squared is equal to one,
and therefore a just equals one.
So now we integrate this with respect to u,
so we're going to have one half,
and then we'll have u times the square root
of the quantity u squared plus one,
plus a squared, of course, is just one,
so we have natural log, absolute value of u,
plus the square root of the quantity
u squared plus one,
so all of this is the anti-derivative,
and we're integrating with respect to u
from zero to one.
So now we have one half times the integral from zero to pi,
and now we're getting ready to substitute
one for u and then zero for u.
So when u is equal to one, notice how here
we're just going to have one times
the square root of the quantity one squared plus one.
That's the square root of two.
And then plus natural log absolute value of one,
plus, this is going to be again
the square root of one squared plus one,
the square root of two.
So this is the value when u equals one,
minus when u is zero notice how
we're going to have zero here,
so we'll have zero plus natural log,
and again, if u is zero we have zero
plus the square root of one, which is just one,
so natural log one.
And notice how natural log one is equal to zero,
so this simplifies out.
Notice that all this is just a constant,
so let's write this as one half
times the quantity square root two
plus natural log absolute value of
one plus square root two,
all of this times integral from zero to pi
of dv, if we want one dv.
And then we integrate this back to v.
Of course, anti-derivative of one is...
Well, the anti-derivative of one with respect to v
would just be v.
So this gives us one half times the quantity
square root two plus natural log
absolute value of one plus square root two
times, well of course when v is pi we have pi.
When v is zero we have zero.
So the exact value of the surface area
is equal to one half pi times the quantity
square root two plus natural log.
We actually don't need the absolute value here
because that's positive,
so natural log of one plus square root two.
Let's also get a decimal approximation
for this surface area.
So, the approximate value is
three point six zero five nine.
And this is surface area,
so the units would be square units
here for the approximate value,
as well as here for the exact surface area,
which is the surface area of this surface here.
I hope you found this helpful.