Arc Length.....&Area of surface....Thanks so much!

[nikchic5]

Ok here is the question...I am studying for a test that I have in a few hours so any help would be reayy appreciated!!

1. Set up the intergral that represents the arc length of the curve y=1/x over the interval [0,3]. Please help

And one more question if anyone could help me with this too!
2. Set up the intergral that represents the area of the surface formes by revolving the graph f(x)=81.x^2 on the interval [0,9] about the y-axis.


Thanks so much for all of you help!!

 

[galactus]

The arc length formula is rather straight forward.

\(\displaystyle \L\\\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx\)

 

[nikchic5]

Is this correct??

S0 is the answer S= (intergral from 1 to 3) (sq.rt 1+1/x^4) dx
Is that correct? Thanks so much..


Also would you help me with the other one please?

 

[tkhunny]

Perimeter of Circle = 2*pi*r

Really, that's about all you need.

 

[galactus]

The formula should be in your calc book. Look it up.

Nothing fancy here. You have f(x) and the limits of integration.

 

[nikchic5]

I am at
2pi (intergral from 0 to 9) of 81-x^2 (sq.rt 1+4x^2) dx

I am drawing a blank...how do I integrate the sq.rt function? And am I doing it riht so far?

Thanks so much!

 

[stapel]

I am at
2pi (intergral from 0 to 9) of 81-x^2 (sq.rt 1+4x^2) dx

Do you mean the following...?

. . . . .(2 pi) int [0,9] [ 81 - x² sqrt(1 + 4x²) ] dx

In other words, the x² is multiplied on the square root. Or do you mean something else?

Thank you.

Eliz.

 

[nikchic5]

Yes

No that is what I am at. Now where do I go from there? The final answer is
2pi int[0,9] x [sq.rt 1+4x^2] dx and I don't see what they get that?
Please help.

 

[galactus]

Find the derivative of 81x^2 and square it; just as in the arc length formula. . It is not 4x^2.

EDIT: now you have 81-x^2. Is it that or 81x^2?.

 

[stapel]

No that is what I am at.

I'm not sure what you mean...? Are you saying that, "no", what I'd posted what not what you'd meant...? If so, what did you mean?

The final answer is 2pi int[0,9] x [sq.rt 1+4x^2] dx and I don't see what they get that?

The "final answer" for which of the two integrals? How far have you gotten between the original exercise and wherever you are? (Please show all of your steps and reasoning.)

Thank you!

Eliz.

 

[nikchic5]

Follow up question..

It is 81-x^2 and then you get -2x^2 correct? And then when you sqaure that dont you get 4x^2? I am losing my mind!! Sorry...where am I going wrong? Thanks

 

[galactus]

If it is 81-x^2, then you're correct

f'(x)=-2x; square that and get 4x^2