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Math Help - Volume of a solid

  1. #1
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    Volume of a solid

    I have this problem that wants me to find the volume of a solid. The problem is as follows:

    Find the volume of the solid obtained by rotating the region bounded by y = 0, y = (tan(x))^3sec(x), and x = 1/3*pi, about the x-axis.

    I'm having trouble getting started with this problem. I've never done a problem like this and I can't find anything helpful in my textbook (I'm probably not looking in the right direction). I was wondering if anyone could give me some tips on how to get started or how to approach the problem.

    Thanks for you consideration,

    Austin Martin
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by auslmar View Post
    I have this problem that wants me to find the volume of a solid. The problem is as follows:

    Find the volume of the solid obtained by rotating the region bounded by y = 0, y = (tan(x))^3sec(x), and x = 1/3*pi, about the x-axis.

    I'm having trouble getting started with this problem. I've never done a problem like this and I can't find anything helpful in my textbook (I'm probably not looking in the right direction). I was wondering if anyone could give me some tips on how to get started or how to approach the problem.

    Thanks for you consideration,

    Austin Martin
    Use the formula for volume about the x-axis

    V_{\text{revolution}}=\int_a^{b}f(x)^2dx

    So we would have assuming the left bound is the x-axis

    \int_0^{\frac{\pi}{3}}\tan^6(x)\sec^2(x)dx

    Then let \varphi=\tan(x)
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    Use the formula for volume about the x-axis

    V_{\text{revolution}}=\int_a^{b}f(x)^2dx

    So we would have assuming the left bound is the x-axis

    \int_0^{\frac{\pi}{3}}\tan^6(x)\sec^2(x)dx

    Then let \varphi=\tan(x)
    Thanks for the help.

    I do have a follow-up question, though.

    I applied the formula you suggested and came up with

    \int_0^\frac{\pi}{3}tan^6(x)sec^2(x) = \frac{tan^7(\frac{\pi}{3})}{7} which is not the correct answer. Either I messed up integrating or the previously suggest approach is incorrect. Thanks in advance.
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  4. #4
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    Hello,

    Quote Originally Posted by auslmar View Post
    Thanks for the help.

    I do have a follow-up question, though.

    I applied the formula you suggested and came up with

    \int_0^\frac{\pi}{3}tan^6(x)sec^2(x) = \frac{tan^7(\frac{\pi}{3})}{7} which is not the correct answer. Either I messed up integrating or the previously suggest approach is incorrect. Thanks in advance.
    \tan \frac{\pi}{3}=\frac{\sin \frac{\pi}{3}}{\cos \frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac 12}=\sqrt{3}

    \frac{\tan^7 (\frac{\pi}{3})}{7}=\frac{(\sqrt{3})^7}{7}=\frac{3 \cdot 3 \cdot 3 \cdot \sqrt{3}}{7}=\frac{27 \sqrt{3}}{7}

    Is it the solution ?
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  5. #5
    MHF Contributor kalagota's Avatar
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    multiply the integral by \pi

    V = \pi \int_a^b (f(x))^2 \, dx
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  6. #6
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    Quote Originally Posted by kalagota View Post
    multiply the integral by \pi

    V = \pi \int_a^b (f(x))^2 \, dx
    Yes, this is the solution exactly! Thanks for the help.

    I hate to be a bother, but I do want to make sure I'm doing these problems right, so I have a similar problem:

    "Find the volume of the solid obtained by rotating the region bounded by y = 0, y = e^(5*x), x = 1, and x = ln(3)"

    Applying the same idea before would the volume be:

    \pi\int_0^1e^(10x)dx ?

    Or since it's revolved about the y-axis instead of the x-axis would the upper and lower boundaries be different? Thanks again in advance.
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  7. #7
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    Quote Originally Posted by auslmar View Post
    Yes, this is the solution exactly! Thanks for the help.

    I hate to be a bother, but I do want to make sure I'm doing these problems right, so I have a similar problem:

    "Find the volume of the solid obtained by rotating the region bounded by y = 0, y = e^(5*x), x = 1, and x = ln(3)"

    Applying the same idea before would the volume be:

    \pi\int_0^1e^(10x)dx ?

    Or since it's revolved about the y-axis instead of the x-axis would the upper and lower boundaries be different? Thanks again in advance.
    The part is bold is very vague, you forgot to mention what axis your rotating about. Assuming it is the x-axis and I have read the problem correctly the volume is given by \pi \int_{1}^{\ln 3} e^{10 x} \  dx.

    About the y axis it is a bit more complicated if your sure it's the y-axis I'll post some details on how to do it.

    Bobak
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  8. #8
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    Quote Originally Posted by bobak View Post
    The part is bold is very vague, you forgot to mention what axis your rotating about. Assuming it is the x-axis and I have read the problem correctly the volume is given by \pi \int_{1}^{\ln 3} e^{10 x} \  dx.

    About the y axis it is a bit more complicated if your sure it's the y-axis I'll post some details on how to do it.

    Bobak
    Ah yes, I do apologize for not explicitly stating that it was about the y-axis. I greatly appreciate your help.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by auslmar View Post
    Ah yes, I do apologize for not explicitly stating that it was about the y-axis. I greatly appreciate your help.
    ok no forgetting pi this time "::grumbles::"

    Ok the formula is

    {\color{red}{2\pi}}\int_a^{b}f(x)h(x)

    Where h(x) is the distance from any point in your shaded region that you are attempting to find the volume of to your axis of revolution..in this case it is h(x)=x

    I believe
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  10. #10
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    Yes! Thanks to everyone for the help. The integral is exactly:

    2\pi\int_1^{ln3}xe^{5x}dx
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by auslmar View Post
    Yes! Thanks to everyone for the help. The integral is exactly:

    2\pi\int_1^{ln3}xe^{5x}dx
    Do you need help evaluating it?
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  12. #12
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    Quote Originally Posted by Mathstud28 View Post
    Do you need help evaluating it?
    Oh no, I'm sorry for not mentioning it. I used integration by parts. Thanks again for the help.
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