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Jabatheblob1

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Joined: 2015-03-01

Posts: 856

More Maths Stuff

So i need to find the surface area of the function revolved around the x-axis
2sin(x) + 3
between 0-1.4
How do this? I keep getting above the actual answer. The answer should be below 45 and i am getting 60 or more.

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Kaslai

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From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: More Maths Stuff

So i need to find the surface area of the function revolved around the x-axis
2sin(x) + 3
between 0-1.4
How do this? I keep getting above the actual answer. The answer should be below 45 and i am getting 60 or more.

The answer is indeed below 45 for surface area

All you need to do to find the surface area is integrate the circumference of the circle over the distance you need.

The circumference for a circle is of course pi * diameter. You have a function that describes the radius, so all you need to do to obtain the diameter is to multiply that by two. You can then integrate that circumference over the distance you need.

I highly suggest you Google your questions. This sort of question is not really too hard to find the answer to online with just a quick search.

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: More Maths Stuff

I highly suggest you Google your questions. This sort of question is not really too hard to find the answer to online with just a quick search.

I've already done that. I spent two hits trying to figure out what I did wrong. I am not the best at U-Substitution with anti derivatives of sin functions so I can't do it on paper. The assignment actually requires us not to do it on paper. It's a technology assignment. And it's not as simple as plugging in the circumference formula since there are infinite radius. I used the arc length formula, sqrt(1+(dy/dx)^2). And if the radius was constant I would have been able to put 2pi*r on the outside of the integral. Since r is ever changing, I replace it with the function and put it back in the integral. In the end I get something along the lines of:
2pi*intergeal from 0-1.4 of [(2sin(X)+3))*sqrt(1+2cos^2(X))]
Then solve for the domain. Yet, that doesn't give me the answer I am looking for. And I use online integral calculators(it's not the online calculator that's wrong, I've used 3 different sources and a ti-84 and still got the same answers)

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Kaslai

Official Caroler

From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: More Maths Stuff

since there are infinite radius.

I can only assume you mean that there are infinite radii, but the size of the radius is bounded by the function you provided. The point of integration is to make the "infinite" radii give you a useful result.

I think you're just doing the integral wrong. You can integrate each side of an addition operation separately, so you essentially have the integral of 2sin(x) and the integral of 3 added together. The integral of 2sin(x) probably isn't what you think it is, either.

Information about integrating trig functions

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: More Maths Stuff

since there are infinite radius.

I can only assume you mean that there are infinite radii, but the size of the radius is bounded by the function you provided. The point of integration is to make the "infinite" radii give you a useful result.

I think you're just doing the integral wrong. You can integrate each side of an addition operation separately, so you essentially have the integral of 2sin(x) and the integral of 3 added together. The integral of 2sin(x) probably isn't what you think it is, either.

Information about integrating trig functions

I don't think you understand what I am asking. It's not as simple as breaking up a function like that, let alone the sin(X) is not the problem. I understand the antiderivitave of individual terms but finding the antiderivitave of functions that require the chain or prduct rule, which is present in this scenario. Unless you can show me how to do this step by step, providing links to websites won't be useful to me :/

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Kaslai

Official Caroler

From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: More Maths Stuff

Find the indefinite integral of 2sin(x) and call it a(x)
Find the indefinite integral of 3 and call it b(x)
The indefinite integral of 2 * pi * (2sin(x) + 3) is 2 * pi * (a(x)+b(x))

Your answer should be the result of evaluating the indefinite integral over the domain you have specified.

Jabatheblob1

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Joined: 2015-03-01

Posts: 856

Re: More Maths Stuff

Find the indefinite integral of 2sin(x) and call it a(x)
Find the indefinite integral of 3 and call it b(x)
The indefinite integral of 2 * pi * (2sin(x) + 3) is 2 * pi * (a(x)+b(x))

Your answer should be the result of evaluating the indefinite integral over the domain you have specified.

That's only half of the intergeal I wrote. The overall part in the integral requires the product of the original function and the arc length of the function. Which the integral of the arc length between 0-1.4 alone requires U-Sub due to the cos^2. This is why he had us use tech

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Kaslai

Official Caroler

From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: More Maths Stuff

Alright I did a little more looking and you're right you need to use the arc length formula. The answer I got was 62.8.

The only thing that I see wrong with your integral is that you forgot to square the two in the arc length formula. (The two that you brought in along with the 2sin(x))

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: More Maths Stuff

Alright I did a little more looking and you're right you need to use the arc length formula. The answer I got was 62.8.

The only thing that I see wrong with your integral is that you forgot to square the two in the arc length formula. (The two that you brought in along with the 2sin(x))

Yeah I assumed I typed something wrong. Regardless, it's still not within the range of a reasonable answer which is the problem I am having. I get the answer you got, when the answer should at least be less than 45

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Kaslai

Official Caroler

From: SEAͩT̓͑TLͯͥͧͪ̽ͧE͑̚

Joined: 2015-02-17

Posts: 787

Re: More Maths Stuff

Alright I did a little more looking and you're right you need to use the arc length formula. The answer I got was 62.8.

The only thing that I see wrong with your integral is that you forgot to square the two in the arc length formula. (The two that you brought in along with the 2sin(x))

Yeah I assumed I typed something wrong. Regardless, it's still not within the range of a reasonable answer which is the problem I am having. I get the answer you got, when the answer should at least be less than 45

But that is the answer. I have verified it as much as I possibly can. Perhaps the book is wrong? It wouldn't be the first time that's happened.

Jabatheblob1

Member

Joined: 2015-03-01

Posts: 856

Re: More Maths Stuff

That is the answer. I have verified it as much as I possibly can. Perhaps the book is wrong? It wouldn't be the first time that's happened.

Ok i found the problem/confusion.
When i was doing this in class with all my classmates, we all reached the same problem where the answer had to be under 45. This information was provided by our teacher who i guess wasn't paying attention. Since this was a technology assignment, he was being smug and only gave us that information, and told us to figure out the answer online. I emailed him and he realized that he was far off. The answer had to be under 150. I honestly don't understand how that got so misinterpreted by him to us, but we spent all of class, plus out of class time figuring out what in the hell we did wrong.

Sorry Kaslai if i got annoyed and seemed rude, I've been really frustrated with this problem all day.

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