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1) Find the arc length for the following curvesy2 = 4(x + 4)3 X= C+- 4y20 <x <2 1<y<22) Find the surface area resulting from the rotation of the curve a...

Question

1) Find the arc length for the following curvesy2 = 4(x + 4)3 X= C+- 4y20 <x <2 1<y<22) Find the surface area resulting from the rotation of the curve about X axis9x = y2 + 18, y=Vi+Ax,2 <* <6 1 <*<53) Find the surface area resulting from the rotation of the curve 0 <*<1 about th Y axis_V=1-x,

1) Find the arc length for the following curves y2 = 4(x + 4)3 X= C+- 4y2 0 <x <2 1<y<2 2) Find the surface area resulting from the rotation of the curve about X axis 9x = y2 + 18, y=Vi+Ax, 2 <* <6 1 <*<5 3) Find the surface area resulting from the rotation of the curve 0 <*<1 about th Y axis_ V=1-x,



Answers

The given curve is rotated about the y-axis. Find the area of the resulting surface.

$ x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1 $ , $ 0 \le y \le 1 $

Before plugging into the integral. The first thing we should do is we should different. She ate 1/2 parentheses. Why squared plus two to the 1/2 in order to then be able to plug in to the integral. So we end up with differentiating This meant up with Y to the fourth plus two y Square, which means now we can plug it into our intro from 1 to 2 to pie. Why time squirt of one plus y to the fourth plus two y squared times D Wykes of the variables. Why not acts? Okay, now that we have this, we know that we can pull out the to pile on the outside. We have integral from 1 to 2. Why cute? That's why, again simplifying it. It's best to simply as much as possible for integrating, because now that we're integrating, we're taking the power rule. It's so much easier now that we don't have anything under a square root plug end. We end up with 21 pie divided by two

This question states that the given curve is rotate about the y axis and we know we need to find the area. What we know we need to do is consider the fact that we're looking at bounds between zero and 12. Exit between zero and 12 has given the problem. We also know we have to pile on the outside. Their formula essentially is to pi times the integral from a to b times X squared of one plus f of x f prime of acts which essentially the derivative times d of acts so plugging into this formula. And again, this is listed in the textbook as well. With examples, we know we can simplify. Now what we know we can do is there's a lot going on here with square root. So I would recommend using u substitution. If you is four plus tax, we know acts is you minus four. This helps us because now we know we can write this out. Integrate this. We integrate. Remember, we know what us and we know what X is. We know we integrate by increasing exported by one dividing by the new expert is you can see I'm dividing by the new exploding a five over too. Now you can plug in. Remember, you're always doing your upper bounds minus your lower bounds in this context, and I end up with 3712 pi divided by 15.

This question asked us to find exact area of the surface by rotating the curb about the X axis. As it stated, What we know we need to do is we know we need to first off, figure out the bounds. They've actually given this to us of exes between zero and three. This means zero and three are about. Now we know we pull out to pie because we know we're gonna be rotating it as it's stated in the problem. And then we know that we have Why squared is X plus one. What we know this means is that if axes y squared minus one, then D axe over D. Y is equivalent to why this is critical here. Because now what we know we have is remember when we're writing this were essentially plugging in using X. So that is the same thing. That's why you can consider this to be. And then we know that we need to integrate. So in order to do that, we know we're going to be We know we're going to be using the for the multiplication pattern. Essentially distribution eight times people see, is a B plus a C So it's from 0 to 3 to pie X plus five over four de axe. We know we're going to be integrating this. We use the power rule of increasing the exploded by one dividing by the new exponents. That's how you integrate. And then we plug in and what we end up with is pi. Over six time 17 squared of 17 minus five squared of five.

For this problem we want to find the exact area of the surface obtained by rotating the curve X equals one third Y squared plus two to the power of three over to about the X axis. So to begin, we want to rearrange that equation to get Y as a function of X. Doing so we should get that. Why is equal to three X. To the power of to over three minus two. That means that why prime of X will equal one half of three X. The power of to over three minus two to the power of negative one half times. Now we want the derivative of the inside. So that would be the derivative of three to the power of to over three X times X. To the power of or three to the power to over three times X to the power of to over three. So be 3 to 3 to the power of to over three times to over three X to the power of to over three minus one. So X power of negative 1/3 minus two. That means ah one second here. The most simplified form that we can get that derivative into would be one over three X. To the power 1/3 times the square root of three X. Or 2/3 minus two. Yeah. Which then means that our Y prime squared which will need for our surface area equation, it's going to come out to 1/3 X. The power of to over three times three X to the power of to over three minus two. So now that we have that we can write down our surface area equation which will be the integral from 1 to 2. Actually we do need to be careful because we're trying to get the rotation about the X axis. You need to figure out what X would be when Y equals one and when Y equals two. So X. Of one should come out to be route three. An ex of two come out to be um to route six. So we'll have that our surface area is going to equal the integral. From route three up to two. Route six two pi Y. So two pi times the square root of three acts the power of to over three minus two times the square root of one plus one. Over three acts the power of to over three times three acts the power of to over three minus two. The X. So having that, we should now be able to get the indefinite expression for our uh integral first of all. One moment. Whoa. I frantically type offscreen here should take just a second. All right. Unfortunately, the most simplified form, that integral is still rather ugly. It's going to be let's see here, negative pi over six Times three to the power of five over I'm going to zoom in here. Three to the power of 5/6 times the cubed root of x minus three. Route three X times the square root of three to the power of two or three acts to the power of to over three minus two times the square root. Now this is the big one of negative three times three to the power of to over three. So that's negative three. Um students uh times three to the power of to over three would be negative three to the power of 5/3. Yeah, times X to the power of to over three minus three. To the power of 1/3 over X. The power of to over three plus six, all divided by two minus three X. To the power of to over three. I said I know that's pretty gruesome but we worked through it there and we're evaluating that from Route three up to two. Route six. Doing so. That should evaluate to drum roll, please? 21 pi over two.


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