2 Problems, Finding exact surface area (integration length of functions)

[Creager]

1. y=e^-x if this infinite curve is rotated about the x-axis. find the area of the resulting surface

SA = integral(0,infinity) (2*pi*y*sqrt(1+(dy/dx)^2)

I am stuck at 2pi*integral(e^-xsqrt(1+(-e^-x)^2)

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2. find exact surface area by rotating about the y-axis x=acosh(y/a) -a<y<a

I have no idea how to work this one at all. rotating about y axis is difficult as it is... adding hyperbolic and the extra variable as the constant got me..

 

[Deleted member 4993]

1. y=e^-x if this infinite curve is rotated about the x-axis. find the area of the resulting surface

SA = integral(0,infinity) (2*pi*y*sqrt(1+(dy/dx)^2)

I am stuck at 2pi*integral(e^-xsqrt(1+(-e^-x)^2)............ substitute ............. e^-x = tan(Θ)

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2. find exact surface area by rotating about the y-axis x=acosh(y/a) -a<y<a

I have no idea how to work this one at all. rotating about y axis is difficult as it is... adding hyperbolic and the extra variable as the constant got me..

.

 

[DrPhil]

2. find exact surface area by rotating about the y-axis x=acosh(y/a) -a<y<a

I have no idea how to work this one at all. rotating about y axis is difficult as it is... adding hyperbolic and the extra variable as the constant got me..

When you rotate about the y-axis, you make circles of radius x. expand to make a band with a height dy, and the incremental area is

\(\displaystyle dA = 2 \pi\ x(y)\ dy = 2 \pi\ \cosh^{-1}(y/a)\ dy\)

The total area is the integral of dA:

\(\displaystyle \displaystyle A = 2\pi \int_{-a}^{a}\cosh^{-1}(y/a)\ dy = 4\pi \int_{0}^{a}\cosh^{-1}(y/a)\ dy\)

No normal person would know that integral without looking it up! But if you have a reasonable table of integrals, the whole problem shouldn't be too bad.

 

[Creager]

When you rotate about the y-axis, you make circles of radius x. expand to make a band with a height dy, and the incremental area is

\(\displaystyle dA = 2 \pi\ x(y)\ dy = 2 \pi\ \cosh^{-1}(y/a)\ dy\)

The total area is the integral of dA:

\(\displaystyle \displaystyle A = 2\pi \int_{-a}^{a}\cosh^{-1}(y/a)\ dy = 4\pi \int_{0}^{a}\cosh^{-1}(y/a)\ dy\)

No normal person would know that integral without looking it up! But if you have a reasonable table of integrals, the whole problem shouldn't be too bad.

Why did you convert cosh to cosh^-1?

From what I learned (or mislearned) in class, this problem should be somewhat like rotating about x axis. Find the derivative, then find the length by sqrt(1+(dx/dy)^2)

then A = integral(2*pi*cosh(y/a)*sqrt(1+(dx/dy)^2)

maybe i have everything backwards because it is around y axis. who knows anymore.

 

[DrPhil]

Why did you convert cosh to cosh^-1?
Sorry about that .. your original typing of the question had "acosh" which I read as "arccosh" instead of "a×cosh".

From what I learned (or mislearned) in class, this problem should be somewhat like rotating about x axis. Find the derivative, then find the length by sqrt(1+(dx/dy)^2)....OOPS - I forgot that step

then A = integral(2*pi*cosh(y/a)*sqrt(1+(dx/dy)^2) dy,,,,Is there also a factor of a missing?

maybe i have everything backwards because it is around y axis. who knows anymore.

When you rotate about the y-axis, you make circles of radius x, where for this problem x is
\(\displaystyle x(y) = a\ \cosh(y/a)\)

Suppose we make a band with a height dy (because we are going to integrate in the y direction).
The slant height is \(\displaystyle ds = \sqrt{(dx)^2 + (dy)^2}\) To keep \(\displaystyle y\) as the independent variable, pull \(\displaystyle dy\) out of the expression. Then

\(\displaystyle \displaystyle ds = \sqrt{1 + \left(\dfrac{dx}{dy}\right)^2}\ dy = \sqrt{1 + \left[(a/a)\sinh(y/a)\right]^2}\ dy = \sqrt{1 + \sinh^2(y/a)} = \cosh(y/a)\ dy\)

\(\displaystyle \displaystyle dA = 2\pi\ x\ ds = \ \cdot\ \cdot\ \cdot \)