Expert: Alon Mandes - 8/15/2009

Question
QUESTION: The diagram shows the curve y=(e^-.5x) (1+2x)^.5 and its maximum point M. The shaded region between the curve & the axes is denoted by R. Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of pi & e. Thank you

ANSWER: Our curve is f(x)=√(1+2x)*e^(-½x). Let's find point of intersect with x-axis :
√(1+2x)*e^(-½x)=0 --> √(1+2x)=0 --> 1+2x=0 --> x=-½ . Therefore :
  0          0
V=∫πf²(x)dx=π∫(1+2x)e^(-x)dx =
-½        -½
   0          0
V=π∫e^(-x)dx + π∫xe^(-x)dx .
-½          -½

The 1st integration is immediate , & the 2nd can easily be calculated via "integration by
parts" . So, I will leave it to you to continue from here as an exercise. Any trouble,
I'm here .

Alon.

---------- FOLLOW-UP ----------

QUESTION: I still cant get the correct answer, I got this pi(-1+2e^.5). The real answer is this pi(-3+2e^.5). I this some of my sign wrong. I check already still didnt get where I got wrong. Please help me. Many Thanks

Answer
Let's 1st calculate ∫xe^(-x)dx : Let's denote u=x & v'=e^(-x). Therefore :
∫uv'       = uv       -∫u'v
∫xe^(-x)dx = -xe^(-x) - ∫-e^(-x)dx
         = -xe^(-x) + ∫e^(-x)dx
         = -xe^(-x) - e^(-x) = -(x+1)e^(-x) .
Now,
   0          0
V=π∫e^(-x)dx + 2π∫xe^(-x)dx .
-½          -½

= -πe^(-x){from 0 to -½} - π(x+1)e^(-x){from 0 to -½}
= -π--πe^(--½) -2π[1-(-½+1)e^(--½)
= -π+πe^(½)-2π+2*½*e^(½)
= π[-3+e^(½)] = π[-3+√e] .

Alon.