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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 xaxis, 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
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 xaxis, 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0eep 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).

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so whats the question?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\pi \int\limits_{0}^{1} [\sqrt[3]{x}]^2 dx\] ??

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0surface area.... ok I wonder if 2pir would work :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[2 \pi \int\limits_{0}^{1} \sqrt[3]{x} dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or do we need to use the integral for the length of a curve?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0something i found intriguing was that if you gut out a circle so that it becomes a donut; with the Rr = 1 that you are 1 off from the circumference. R=6 and r=5 36pi  25pi = 11pi 2pi6 = 12pi.......
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