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anonymous

  • 5 years ago

Let f(x) = the cube root of x. Find the area of the surface generated by revolving the curve y = f(x) around the x-axis, for x ranging from 0 to 1. Keep in mind that the derivative of the cube root of x tends to infinity as x tends to 0, so the area will be equal to an improper integral. Be sure that you express your answer so that it is an actual real number, not an expression which evaluates to infinity minus infinity. For example, if the indefinite integral were ln(x) + ln(2/x), you would not want to plug in x = 0, and write the answer as ln(0) + ln(2/0). Instead, you would want to si

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  1. anonymous
    • 5 years ago
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    eep in mind that the derivative of the cube root of x tends to infinity as x tends to 0, so the area will be equal to an improper integral. Be sure that you express your answer so that it is an actual real number, not an expression which evaluates to infinity minus infinity. For example, if the indefinite integral were ln(x) + ln(2/x), you would not want to plug in x = 0, and write the answer as ln(0) + ln(2/0). Instead, you would want to simplify the indefinite integral to ln(x times 2/x) = ln(2).

  2. amistre64
    • 5 years ago
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    so whats the question?

  3. amistre64
    • 5 years ago
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    \[\pi \int\limits_{0}^{1} [\sqrt[3]{x}]^2 dx\] ??

  4. amistre64
    • 5 years ago
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    surface area.... ok I wonder if 2pir would work :)

  5. amistre64
    • 5 years ago
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    \[2 \pi \int\limits_{0}^{1} \sqrt[3]{x} dx\]

  6. amistre64
    • 5 years ago
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    or do we need to use the integral for the length of a curve?

  7. amistre64
    • 5 years ago
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    something i found intriguing was that if you gut out a circle so that it becomes a donut; with the R-r = 1 that you are 1 off from the circumference. R=6 and r=5 36pi - 25pi = 11pi 2pi6 = 12pi.......

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