Question

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.

Answer 1

Ok, so how did you want to set this up? Disks?

Answer 2

? disks?

Answer 3

explain all my professor said was

Answer 4

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 simplify the indefinite integral to ln(x times 2/x) = ln(2).

Answer 5

this is all my professor gave me this is homework and i dont understand it

Answer 6

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 simplify the indefinite integral to ln(x times 2/x) = ln(2).

Answer 7

explain all my professor said was

Answer 8

sorry, I thought you said volume, not surface area.

Answer 9

do you know how to do this

Answer 10

??

Answer 11

So how did you set up the integral

Answer 12

Or have you yet.

Answer 13

didnt do it yet but i know this much idk if its right

Answer 14

S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx

Differentiate f(x) = ∛(x) = x^(1/3) to get:

f'(x) = (1/3)x^(-2/3)
= 1/(3x^(2/3))

Answer 15

\[\frac{d}{dx}x^{1/3} = \frac{1}{3\sqrt[3]{x^2}}\]

That's right. So we plug in that for dy/dx in the integral (squaring it) and evaluate it as an improper integral.

Answer 16

~=1.885

Answer 17

can you explain how you got this because i need the work for it so i can try to understand it all

Answer 18

because ive got a final tomrrow and i sorta need to know this all

Answer 19

\[\lim_{b\rightarrow 0} \int_b^1 \sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx \]

Answer 20

so first its
S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx

Differentiate f(x) = ∛(x) = x^(1/3) to get:

f'(x) = (1/3)x^(-2/3)
= 1/(3x^(2/3))

Answer 21

then its the limit as etc..
then its d/dx
and then wa because now i still didnt get the answer

Answer 22

Hrm.. I'm not sure that's right.. Seems too awful

Answer 23

Ah. I forgot you multiply the radical by the original f(x).

It should be
\[\lim_{b\rightarrow 0} \int_b^1 \sqrt[3]{x}\sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx\]

Answer 24

That's much better.

Answer 25

so this is in order

Answer 26

so first its
S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx

Differentiate f(x) = ∛(x) = x^(1/3) to get:

f'(x) = (1/3)x^(-2/3)
= 1/(3x^(2/3))

Answer 27

limb→0∫1bx−√31+(13x2−−−√3)2−−−−−−−−−−−−√dx

Answer 28

then

Answer 29

ddxx1/3=13x2−−√3

Answer 30

this doesnt come outto be the same wouldnt d/dx change since the limit changd?