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Math Help - Very Quick Question Regarding Surface Area of Revolution (Integration)

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    Very Quick Question Regarding Surface Area of Revolution (Integration)

    An artist is designing a wine glass in a flower shape, which can be generated by rotating y= sqrt(x) and x= y, between x= 0 and x= 1, about the x-axis. What is the surface area (which contains both the inside and the outside surfaces) of such a glass?

    Alright, quick question: until now, I've dealt with Surface Integral problems that involved only one function [for instance, y= sqrt(x)], so inputing 'y' into the surface area formula [2 π y ds] hasn't been a problem. BUT, this one involves two [y= x and y= sqrt(x)] functions. So, my question is, how do I set it up, i.e. how do I enter them into the formula?

    Thanks in advance!
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    Quote Originally Posted by bhaktir View Post
    An artist is designing a wine glass in a flower shape, which can be generated by rotating y= sqrt(x) and x= y, between x= 0 and x= 1, about the x-axis. What is the surface area (which contains both the inside and the outside surfaces) of such a glass?

    Alright, quick question: until now, I've dealt with Surface Integral problems that involved only one function [for instance, y= sqrt(x)], so inputing 'y' into the surface area formula [2 π y ds] hasn't been a problem. BUT, this one involves two [y= x and y= sqrt(x)] functions. So, my question is, how do I set it up, i.e. how do I enter them into the formula?

    Thanks in advance!
    The inside surface is a cone, you should know the formula for surface area of a cone. So you really only have to evaluate one surface integral (for the outside).
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    But, how would you set up this problem?

    I'm sorry but I still don't understand this.
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    Quote Originally Posted by bhaktir View Post
    But, how would you set up this problem?

    I'm sorry but I still don't understand this.
    You said you know how to evaluate a surface area when you only have one function. Like I said, you only need to integrate one function (the square root) for the surface area of the outside of the glass, because the inside is a cone, which you should already be able to evaluate without integration.
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    All I know is the formula (2 π y ds). I really don't know the rest. If you could just show me how to set this problem up, I'd be extremely grateful. Thanks!
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    Never mind, I got it.

    For future reference, the cone formula isn't necessary in attaining the answer.

    Thanks for the help!
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    Re: Very Quick Question Regarding Surface Area of Revolution (Integration)

    Quote Originally Posted by bhaktir View Post
    Never mind, I got it.

    For future reference, the cone formula isn't necessary in attaining the answer.

    Thanks for the help!
    It is if you want the inner surface area as well...
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