INSTRUCTOR: Find the surface area of the surface with parameterization r of u comma v equal to angle bracket u plus v comma u squared comma 2v close angle bracket with 0 less than or equal to u less than or equal to 3 and 0 less than or equal to v less than or equal to 2. To set this up, to find the surface area, we first need to find t sub u and t sub v. That is the partial of r with respect to u and the partial of r with respect to v.

So t sub u is equal to angle bracket 1 comma 2u comma 0 angle close angle bracket. And tv is equal to angle bracket 1 comma 0 comma 2. Now, we need this because the definition of our surface area is a double integral of the magnitude of tu crossed with tv over that region.

So tu cross tv-- remember, we can think of this as a determinant of a 3-by-3 matrix, with the first row of i, then j, then k; a second row of 1, then 2u, then 0; and a third row of 1, then 0, then 2. We find the determinant of that. And we find that is equal to 4ui minus 2j minus 2uk.

Now we can find the magnitude of that cross product. Magnitude of tu cross tv is equal to the square root of 16u squared plus 4 plus 4u squared, end square root. So this is the square root of 20u squared plus 4, end square root. Or I'd like to go ahead and write that as-- we'll factor out a square root of 4. So that's 2 square root of 5u squared plus 1, end square root.

Now, for our surface area, our surface area will be equal to the double integral over the region-- let's see. What is our region? We have u ranges from 0 to 3, and v ranges from 0 to 2. So we're going to go ahead and set this up as the integral from 0 to 3 with respect to u of the integral from 0 to 2 with respect to v of the function square root of 20u squared plus 4, end square root.

Now, our first inner integral we can evaluate to say that this is equal to integral from 0 to 3 with respect to u of 2 square root of 20u squared plus 4, end square root. And based on what I said earlier about the magnitude of tu cross tv, this is actually equal to the integral from 0 to 3 of 4 square root of 5u squared plus 1, end square root. And that's with respect to u.

I'm actually going to refer to a table of integrals for this. So I'll go ahead and write this out. But this is actually a fairly common integral to work with. We'd make a trig substitution.

So based on my table of integrals, this is going to be equal to 4 times the evaluation of 1/2 times u times the square root of 5u squared plus 1-- end square root-- plus 1 over 2 square root of 5-- end fraction-- times natural log of parentheses square root of 5 times u square root of 5-- end square root-- times u plus the square root of 5u squared plus 1-- end square root, end parentheses, end bracket, because we're evaluating from 0 to 3. And the 4 is multiplied by that entire evaluation.

Now, this is equal to 6 square roots of 46-- end square root-- plus a fraction 2 denominator of square root of 5-- end square root, end fraction-- times natural log open parentheses 3 square root of 5 plus square root of 46, end square root, end parentheses. And this is approximately 43.02 for our surface area.