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# Surface integral ex2 part 2

Video transcript

Now that we've set up
the parametrization, let's try to evaluate
the integral. And the next thing
we'll do is essentially try to express ds right over
here in terms of du and dv. And we've seen this before. ds is going to be equal to the
magnitude of the cross product of the partial of r with
respect to u crossed with the partial of r
with respect to v du dv. So first, let's take
the cross product, and we'll do that
with a 3 by 3 matrix. I'll do that right over here. We set up, and I'll just kind
of fill in what r sub u and r sub v is in the actual
determinant right over here. So first we have our
components, i, j, and k. And now first, let's
think about what r sub u is, the partial
of r with respect to u. Well, its i component
is going to be 1. The partial of u with
respect to u is just 1, so its i component is 1. Its j component
is going to be 0, partial of v with
respect to u is 0. v does not change
with respect to u, so this is going to be 0. This should be
parentheses around here. Partial of this with respect to
u is just going to be 1 again. The partial of v squared
with respect to u is just 0. So this is just 1 again. And then r sub v, the partial
of r with respect to v, the i component
is going to be 0, j component is going to
be 1, and the partial of u plus v squared with
respect to v is going to be 2v. So it's a pretty
straightforward determinant, so let's try to evaluate it. So the i component,
it's going to be i. So we cross out this
column, this row. It's going to be 0 times
2v minus 1 times 1. So essentially, it's just
going to be negative 1 times i. So we're going to
have negative i, so this is equal to negative i. And then the j
component-- and we're going to have to put a negative
out front, because, remember, we do that checkerboard pattern. So cross out that
row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v
minus 0 times 1. So it's just going to be 2v. But since it was the
j component, which is going to be negative, it's
going to be negative 2vj. Let me make sure
I did that right. That column, that row, 1
times 2v is 2v minus 0 is 2v. The checkerboard pattern,
you'd have a negative j. So you have negative 2vj. And then we have
find the k component. Cross out that row, that
column, 1 times 1 minus 0. So it's going to be plus k. So if we want the
magnitude of this, this whole thing right
over here is just going to be the
square root-- I'm just taking the magnitude
of this part right over here, the
actual cross product. It's going to be
negative 1 squared, which is just 1, plus negative
2v squared, which is 4v squared, plus 1
squared, which is just 1. So this whole thing
is going to evaluate to-- we have 2 plus
2v squared du dv. Or actually, let me-- I
almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want,
maybe it'll help us a little bit if we factor
out a 2 right over here. This is the same thing as 2
times 1 plus 2v squared du dv. If you factor out
the 2, you get this is equal to the square root of
2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to
evaluate the surface integral. So let's do it. All right. So let me just write
this thing down here. So I'm going to write everything
that has to do with v, I'm going to write in purple. So I'm just going to write
the ds part right over here. It is the square root of 2
times the square root of 1 plus 2v squared. And then we're going
to have du, which I'll write in green,
du, and then dv. So this is just the ds part. This is just the ds. Now, we have the
y right over here. y is just equal to v. All right,
so I'll write that in purple. So y is just equal to v. That is
y-- let me make it very clear-- and all of this is ds. And now I can write the
bounds in terms of u and v. And so the u part,
u is the same thing as x. It goes between 0 and 1. And then v, v is
the same thing as y, and y goes between, or
v goes between, 0 and 2. And I now think we're
ready to evaluate. And the u and v variables
are not so mixed up, so we can actually separate
out these two integrals, make this a product of
two single integrals. The first thing, if we look
at it with respect to du, all this stuff in purple is just
a constant with respect to u, so we can take it out
of the du integral. We can take all this purple
stuff out of this du integral. And so this double integral
simplifies to the integral from 0 to 2 of-- I'll write it
as the square root of 2v times the square root of
1 plus 2v squared. So I factored out
all of this stuff. And then you have times the
integral from 0 to 1 du, and then times d,
and then you have dv. And now, if this was really
complicated, I could say, OK, this is just going
to be a function of u. It's a constant
with respect to v. And you could factor
this whole thing out and separate the integrals. But this is even easier. This integral is just
going to evaluate to 1. So this whole thing
just evaluates to 1. And so we've simplified this
into one single integral. So this simplifies
to-- and I could even take the square root of 2
out front-- the square root of 2 times the integral
from v is equal to 0 to v is equal to 2 of v
times the square root of 1 plus 2v squared dv. And so we really are
in the home stretch of evaluating this
surface integral. And here, this is basic. This is actually a little
bit of u substitution that we can do in our head. If you have a function, or
kind of this embedded function right over here,
1 plus 2v squared, what's the derivative
of 1 plus 2v squared? Well, it would be 4v. And we have something
almost 4v here. We can make this 4v by
multiplying it by 4 here, and then dividing
it by 4 out here. This doesn't change the
value of the integral. And so now this part
right over here, it's pretty straightforward
to take the antiderivative. The antiderivative
of this is going to be-- we have this embedded
function's derivative right over here, so we can kind of
just treat it like an x or a v. Take the antiderivative
with respect to this thing right
over here, and we get this is essentially 1 plus
2v squared to the 1/2 power. We increment it by
1, so it's 1 plus 2v squared to the 3/2 power. And then you divide by 3/2,
or you multiply by 2/3, so times 2/3. So this is the
antiderivative of that. And then, of course, you
still have all of this stuff out front-- square
root of 2 over 4 And we are going to
evaluate this from 0 to 2. And actually, just to simplify
it, let me factor out the 2/3. We don't have to worry
about that at 2 and 0. So I'm going to factor out
the 2/3, so times 2 over 3. And actually, this
will cancel out. That becomes a 1. That becomes a 2. And so we are left-- this
stuff over here is 1/6 times. And now if you evaluate this
at 2, you have 2v squared. That's going to be 2
times 4 is 8 plus 1 is 9, 9 to the 3/2 power. So 9 to the 1/2 is 3,
3 to the 3rd is 27. So it's going to be equal to 27. And then minus this
thing evaluated at 0. Well, this evaluated at
0 is just going to be 1. 1 to the 3/2 is
just 1, so minus 1. And so this gives us-- oh,
I almost made a mistake. There should be a square
root of 2 up here, square root of 2 over
6 times 27 minus 1. So drum roll. This gives us the value of
our surface integral is, let's see, 27 minus 1 is 26. So we get 26 times the
square root of 2 over 6. And we can simplify
it a little bit more. Divide the numerator
and denominator by 2, and you get 13 square
roots of 2 over 3. And we are done.