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Numerade Educator



Problem 31 Medium Difficulty

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


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


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Video Transcript

Okay. What we want to dio is we want to find the area, which is really a surface area of one side of what is called a winding no standing or thaw Gunnel, um, on a curve that is represented by why equal to x squared. Um, which is, um, And it goes from zero to, um, excuse from Syria to x 0 to 2. Inclusive, um, And beneath, um, curve on a surface. Okay. Um and, um, there is a very good picture in your book, um, to kind of give you a visual. Um, you could also, if you wanted to google, um, winding wall, it will actually show you. Um, I don't dare try to do it. Draw it. I will. But butcher it, But we're gonna go ahead and walk through how we actually, um um how we actually do this? Um, and the surface is represented by the function f of X. Why equal to X plus the square root of why, Okay. And so that surface area is actually equal to the integral over that curve of F. D s. Um, so it's similar to what you've been doing in this whole section and So the first thing we need to do is to define our parametric equation are of tea, and so that is going to be equal to t I plus t Squared J. So it's gonna be based off of this right here. Okay. And so and then, of course, since X goes from 0 to 2 and X's t then we know that t goes from Syria to two inclusive as well. And so are the derivative of our T. Gives me a function vey of tea, which is gonna be I plus two tea J. And the magnitude of e of t is going to be that square root of one squared close to t squared, which is going to give me the square root of one plus four t squared. And so d of s is equal to that. A magnitude of beauty, which is gonna be this quota one plus four t squared times. T t um And so then, um, we can go ahead and start setting up our integral are integral is gonna go from 0 to 2. Uh, it's gonna be over our function. And of course, um, X is t. Um why is t square So we have the square to t squared times that square root of one place fourty squared TT So this is the integral that were really, um, wanting to integrate. And so let's go ahead and simplify. So this becomes the integral from 0 to 2 of, um, to tee times, the integral times says, Excuse me, the square root of one plus fourty squared t t. And so we're gonna have to do You said, um, you were gonna let you be that one plus four t squared. So d'you is equal to a T T T. And all I have is, um, to t g t So I'm in divide both sides by 1/4 to give me that to t t t. And so now what we're gonna do is change or integral, as can be 1/4 integral. Um uh, you to the 1/2. Do you? And I'm gonna go ahead and change my upper and lower limits. And so if, um, t is zero, then we have, um, use one. If t is two in here, then we have a u of 17. And so this becomes 1/4 times 2/3 you to the three halves, and we're gonna evaluate it at 17 and at one. So this becomes 16? Um, you, uh, not you were gonna evaluate it. So is going to be 17. Raised to three house power minus one.

University of Central Arkansas
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