Expert: Paul Klarreich - 5/6/2006

Question
Hi.
Here's the question again:
{t=inf a[1 - e^(-kt)]e^(-bt).dt
N| -------------
}t=0 k
Solve the improper integral to show that S=Na/[b(b+k)]

I got stuck on another question:
a) Find the volume of the solid of revolution formed by rotating about the x axis the area bounded by y=(2e^x)-2, y=0,x=0 and x=2. Answer in exact and decimal

b)The function defined by y=(b/a)square root of (a^2-x^2) has its graph a semi ellipse. When the semi ellipse is rotated about the x axis, an ellipsoid results.
i) find the formula for the volume of an ellipsoid
ii)use the previous result to write down the volume of a sphere with radius r.

Thank you

Answer
Hi, Emma,
 
You wrote:
Here's the question again:
{t=inf
N|      a[1 - e^(-kt)]e^(-bt) dt
}t=0 k
Solve the improper integral to show that S=Na/[b(b+k)]
--------------------
You mean to EVALUATE the integral.  Try this:

 {t=inf
aN|      [e^(-bt) - e^(-kt - bt)] dt
 }t=0 k

Remember:
First:  e^p e^q = e^(p+q)
Second: the integral of  e^pt dt = 1/p e^pt, so:

The integral gives us:
   e^-bt   e^(-kt - bt)
aN[ ----- - ----------- ]
    -b       -k - b

Lots of minus signs there, so:

   e^-bt   e^(-kt - bt)
aN[ ----- + ----------- ]
    -b       k + b

Now, then, as  t -> inf, e^-(anything)t  --> 0.  (that's good -- we can ignore the right hand boundary)

And when t = 0, the left boundary, we have: (don't forget that the left boundary is subtracted)

     1      1
-aN[ --- + ----- ]
    -b    b + k

   1      1
aN[--- - ----- ]
   b    b + k
  b + k - b
aN ---------
   b(b + k)

      k
aN ---------
   b(b + k)

Hmmmmm...  Did we lose a k, somewhere?

---------------------------
I got stuck on another question:

a) Find the volume of the solid of revolution formed by rotating about the x axis the area bounded by y=(2e^x)-2, y=0,x=0 and x=2. Answer in exact and decimal

Use the method of disks.  Each disk has as its radius the value of y = 2e^x - 2, and has dx as its thickness.  It's a cylinder, volume equal to:

V = pi r^2 h
dV = pi (2e^x - 2)^2 dx
dV = 4 pi (e^x - 1)^2 dx
dV = 4 pi [e^2x - 2e^x + 1] dx

Integral is

V = 4 pi[e^2x / 2 - 2e^x + x], from x = 0 to x = 2.

V = 4pi[e^4/2 - 2e^2 + 2]

You can do the calculator work now.  I think it's around 51.
 
-----------------------

b)The function defined by y=(b/a)square root of (a^2-x^2) has its graph a semi ellipse. When the semi ellipse is rotated about the x axis, an ellipsoid results.
i) find the formula for the volume of an ellipsoid


Let's just do this in the first quadrant.  Later we'll double the answer to take care of the left side.

V = pi r^2 h, as above.

y = (b/a)sqrt(a^2-x^2)

dV = pi (b^2/a^2) (a^2 - x^2) dx

V = pi (b^2/a^2)[a^2x - x^3/3]  from x=0 to x=a

V =  pi (b^2/a^2)[a^3 - a^3/3]

V =  pi (b^2/a^2)[2a^3/3]

V =  pi (b^2)[2a/3]

V =  2/3  pi ab^2

Now we double it and get

V = 4/3 pi ab^2

That's your formula.

ii)use the previous result to write down the volume of a sphere with radius r.

Easy.  Just set  a = r and b = r.  You get:

V = 4/3 pi r^3