surface area problem (1 Viewer)

[xanny]

Hi, I've been struggling with this question and hope that you guys will find a solution. Thanks alot!

y=sec(x) & y=tan(x), x=0 to x=pi/2 are the limits and revolved around the x-axis.

~show that the inner and outer surfaces of the solid have infinite area.

 

[pLuvia]

y=secx
y=tanx

A = ∫[0 to pi/2] (secx - tanx) dx
= {[ln(secx+tanx)] - [ln(secx)]}0 to pi/2
= (ln ∞ + ∞) - (0 - 1)
= ∞

Hence the inner and outer surfaces have infinite area

I think that's how you do it

 

[xanny]

i think you have to use the formula for the surface area which is 2pi ∫ x sqrt(1+(dx/dy)^2)dy
i am assuming what you did was to find the area between the 2 curve
and what is this "inner and outer surfaces"? that's the part i can't understand.

 

[pLuvia]

Err, I don't know that one sorry

My one is wrong sorry

 

[Stan..]

i think you have to use the formula for the surface area which is 2pi ∫ x sqrt(1+(dx/dy)^2)dy
i am assuming what you did was to find the area between the 2 curve
and what is this "inner and outer surfaces"? that's the part i can't understand.

The Volume of the Solid will allow you to consider both of those in one equasion. Find that, if it is void then your original conjecture is correct.

 

[Stan..]

y=secx
y=tanx

A = ∫[0 to pi/2] (secx - tanx) dx
= {[ln(secx+tanx)] - [ln(secx)]}0 to pi/2
= (ln ∞ + ∞) - (0 - 1)
= ∞

Hence the inner and outer surfaces have infinite area

I think that's how you do it

How does that work? That is the 2d Area between the curves.