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Find the area of the surface generated when the given curve is revolved about the given axis. y= # (e 5k + 5x) for 2sxs2; about the X-axisThe : surface area is squa...

Question

Find the area of the surface generated when the given curve is revolved about the given axis. y= # (e 5k + 5x) for 2sxs2; about the X-axisThe : surface area is square units_ (Type an exact answer; using K as needed )

Find the area of the surface generated when the given curve is revolved about the given axis. y= # (e 5k + 5x) for 2sxs2; about the X-axis The : surface area is square units_ (Type an exact answer; using K as needed )



Answers

Computing surface areas Find the area of the surface generated when the given curve is revolved about the $x$ -axis. $$y=\sqrt{4 x+6} \text { on }[0,5]$$

This problem. We're being asked to find the surface area that is created by rotating a function around the X axis. The picture on the right is a generic example. The problem that we're being given in this problem is our function the square root of five X minus X squared on the region from 1 to 4. To find the surface area of a revolution, we're going to need to use a formula for that be integral from A to B. As written at the top left corner of two pi times are function f of x times the square root of one plus f prime of X quantity squared. You know, our function. We know the bounds A and B, but we need to know the derivative so are derivative of our function in this case should be 1/2 times five x minus X squared, the negative 1/2 times change Rule five minus two acts. We're gonna take all that. We're gonna plug it into our surface area formula surface area. In this case is gonna be the in a girl from our lower bound of one to our upper bound of four of two pi times our function, which is the square root of five x minus X squared times the square root, one plus hour derivative square and are derivative. Every found was 1/2 five acts minus X squared that negative 1/2 times five minus two acts close that bracket, square it inside the square root and the d X At the end of our integral we evaluate that integral. Using a numeric calculator, we find our answer is 15 hi.

All right, because we're rotating around the Y axis. Our surface area formula looks like follows with thanks as our radius square this derivative anything great against ey. All right. No, um, we have why, in terms of X, but we need X in terms of why. Fortunately, this is just a couple steps of algebra, and we get 1/4. Why? Plus 1/4. We already have the bounds for why it ranges from three to 15. And also, since there's ah, since this is a line, we know that it doesn't cross over itself, which is important because we couldn't just use one integral, You know, if the if, why did something, like, you know, if we had something like, um, actually Okay, so we're rotating around. Why? So that wouldn't be a problem if we had something like this. We were rotating around, and it kind of overlapped itself, you know? Then we kind of need a We need a breakdown. The inter girls a little differently, but we just have a line. Okay, so there's no overlap. You know, we can return to the main problem. Fortunately, the derivative dx dy Why is gonna be pretty straightforward again. This is a line. So dx dy y is just equal to 1/4. Okay. And by the way, one plus 1/4 squared is just Route 17/16. Okay? And the reason is the first term, the one is 16/16 in the second term is 1/16. So we get 17/16 and that's just rude. 17/4. Okay. Um, So what this means is that our surface area formula, in fact, I'll just swap it out here, ends up being 1/4. Why? Plus 1/4 times? Um, I don't need to go top and bottom. Just Route 17 over four. See why? And we're going from 3 to 15. Good. No, to make your problem tad easier. We can sort of bring this constant out, all right? And we get if I times Route 17 over to and we're gonna know and then we are going to do well, we basically air just integrating this part, right? So I'm going to kind of jump ahead because this is just an application of the power rule, but we should end up with is in a greeting. Why squared times one a plus 1/4 times. Why? Evaluating from 3 to 15? What you should get doing this out is well, you end up with 1/8 times 15 square. That's to 25 here unless 1/4 times times 15 1/8 times. Nine. Okay, and there's a little bit of number crunching, but basically, at the end of the day, this is 30 around 17. It's not very good. 17 will fix that up to equals 15 pie times, Route 17 and that is your final answer.

Either. Today we're going to look at the curved surface area we get when we rotate the curve. The square to four x plus six about the X axis from X equals zero toe X equals five. No using the surface area formula, we see that we need to use the first derivative of every thanks Now F of X is for X plus six to the half. This makes it easier to use the chain role. So f dash of X. We bring down the half and then we subtract one from our power for X plus six to minus 1/2. And then we multiply by the derivative of what's inside, which is four. So you have to over four X plus six square root. No, in fact, we need to deal with the square of this. So, therefore, over four X plus six. Now we complete this into our surface area formula. The surface area is two pi times the integral from 0 to 5 of F of x, which is four x plus six square root times, this one plus the square of F dash, which is 4/4 x plus six d X now we can bring in our squirt for X plus six. So you have ah, two pi times in school from not five. So I have four X plus six and then for explosive six will cancel with the denominator for X plus six. So we just get plus four dx So this Oops. So this is equal to two pi integral from zero 25 for X plus 10 Square it DX. Now we can use a U substitution. For this we set you you could have four X plus 10 Do you is forex No. Four DX Thandi You ever four his DX. We can use that as a substitution and we need to look at the limits. So when you well, when x zero you is 10 when x is five you is five times four, which is 20 plus 10 is 30 how we can use our formula again. So we pull out the quarter that will get from changing DX to do you ever four. So you have two pi over four intro on new limits 10 2 30 of you to the half Do you Now we add one to the power and divide by the new power. They have pi over two times. 2/3. Well, and this is you 23 of to from 10. 2 30 So I twos council and you to the three of the two is the same of you route to you. So this is just equal to pie. The three times 30 Route 30 minus 10 Route 10. We have a final answer, I think.

In this problem looking to find the surface area generated by revolving the function shown on the right like was eight square root of X around the X access. Uh, this region will be bounded between the expellees nine and 20. First up here is recognizing be need for our surface area formula for revolutions of functions. In this case, in terms of X, we're looking at the interval from a to B of two pi times f of x times the square root of one plus f prime of X squared. In this case, our function is it X and our derivative function is for X to the negative 1/2. So if you plug goes in to our integral, we're looking at the integral from 9 to 20 of two times pi times eight route x times a square of one, plus the derivative that we found a moment ago Square. We'll simplify this down. We can factor out not just the two pi, but the two pi times eight's. Let's 16 high times the integral from 9 to 20 of the square root of X times one plus Here are square vert. Well, uh, be reduced with our squares with one plus 16 over X dx Ah, continuing to simplify 9 to 20 you could distribute the spirit of X inside and we get X Plus 16. The X now each taken anti derivative and evaluations. We had 16 high. The anti derivative is 2/3 next, plus 16 to the three halves evaluated from nine 2 20 which is 16 hi times. 2/3 of 20 plus 16 to the three halves minus 2/3 of nine. Plus 16 the three house. When we simplify this all down, we end up with 2012 3rd Hi.


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