## 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

1. anonymous

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

so whats the question?

3. amistre64

$\pi \int\limits_{0}^{1} [\sqrt[3]{x}]^2 dx$ ??

4. amistre64

surface area.... ok I wonder if 2pir would work :)

5. amistre64

$2 \pi \int\limits_{0}^{1} \sqrt[3]{x} dx$

6. amistre64

or do we need to use the integral for the length of a curve?

7. amistre64

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.......