Find the area of one side of the "winding wall" standing perpendicularly on the curve $y=x^{2}, 0 \leq x \leq 2,$ and beneath the curve on the surface $f(x, y)=x+\sqrt{y}$.

$\frac{1}{6}\left(17^{3 / 2}-1\right)$

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Okay, folks. So in this video, we're gonna take a look at this line integral, which is problem number 31. Without further ado, let's Ah, let's get started with Integral. First of all, this problem is about finding the area of one side of the of the wall standing perpendicular early on the curve. And that area is really just that London to grow because that's what area is you. Uh, you integrate a function along a curve, and that is what an area is. So this problem, even though it might look like a word problem, it's really not. It's really just a simple regular line. Integral problem. Just like any other problem. Okay, so we haven't integral long sea of f x Y Yes. Okay, basically, we did. We have, You know, we have this curve and you should imagine, like a wall standing perpendicular, lee out of the page, facing you, not facing you, but directed towards you. And you're calculating the area, um, of one side of the wall. Okay, that's that's really the geometrical interpretation of this problem. And that's plugging the numbers we have for the function f We have X plus route Why? For DS, I'm gonna, you know, do a little trick here, Um, and write this as two x one plus two x squared the X. Okay. All right. Just used this thing. DS equals one. Plus, why Prime squared the X. All right, um, and this I'm going to simplify. Oh, also, the integration limits I forgot is that that's just zero into. Okay? And this is, um, X plus route. Why? But why is X squared? So I'm gonna have X here, um, multiplied by one plus four x squared dx. And this is really just two X. I mean, integral to eggs one plus four x squared the X between zero and two. Now, if you do a little bit of U substitution, which is ah, a technique where you define your variable in terms of X, something defined you as one plus four x squared. So d you over four is eight x d x over four, which is do x, t x. And if you look at here, you have two x t x right here. Um so narrow. I am going to put in you. So we have route you do you over four because that's what two x dx is it, son? Do you over for my writing is a little bit missy do you wear for All right, You know, if you crank out the algebra here, this is gonna be a lot of algebra. Not really. Um, by you will figure out that time that integration limits is one in 17. And now, if you crank out some more algebra, you're gonna have 1/4 times you the rehab over three. Half between one and 17. No. Now is the time where you ah, do your algebra One stuff hand, Um, and just simply plugging in 17 here in plucking one there and you subtract some things and you add them up, You're going to end up with 1/6. 17 to the power of 3/2 minus one. I skipped through a few trivial and unnecessary intermediate steps. I hope you have the time to, uh, fill those steps in yourself. But its religious algebra with them without much technicality, I guess. And so that's it. Finished video. We have our answer right here. 1/6 times this chunk of numbers. That's it for this video. Thank you for watching. Bye bye.

University of California, Berkeley