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Trig Substitution and Volume of Solids

  • Thread starter Luscinia
  • Start date
  • #1
17
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2 Questions here! (I'm not exactly sure if it's allowed, but I want to avoid posting too many threads)

Homework Statement


1)Find the following definite integrals by using a trigonometric substitution:
d)1/21dx/(√(2x-x2)


Homework Equations


x=asinu


The Attempt at a Solution


From a previous question, I found that the indefinite integral of ∫dx/(√(2x-x2) was arcsin(x-1) + C
Using FTC part 2, I got:
arcsin(1-1)-arcsin(1/2-1)=arcsin0-arcsin(-1/2)=0- -∏/6=∏/6

The answer the teacher gave me was ∏/2
I'm have no idea where I have made a mistake.


Homework Statement


Find the volumes of the solids generated by revolving the region bounded by the curves y=x+2 and y=x2 about (a) the line x=2 (b) the line y=4


Homework Equations


Shell:V= 2∏abxhdx
Disc: V=∏abr22-r12dx


The Attempt at a Solution


a)
x+2=x2
0=x2-x+2
(-b±√(b2-4ac))/2a=(1±√(1-4(1)(-2))/2(1)=(1±3)/2
x=-1 and x=2

2∏-12x*(x+2-x2)dx
2∏(1/3*x3+x2-1/4x4|-12
2∏(8/3-5/12)=

9∏/2
The correct answer was supposed to be 27∏/2


b)
ab(4-(x2)2(4-(x+2))2dx
ab12-9x2+4x-x4dx
∏(12x-3x3+2x[sup2]-1/5x[sup5])|-12
∏[(24-24+8-32/5)-(-12+3+2+1/5)]=
42∏/5
The answer is supposed to be 108∏/5


Thanks!
 

Answers and Replies

  • #2
798
1
For d) It is not
[tex]arcsin(x-1)[/tex]
 
  • #3
I like Serena
Homework Helper
6,577
176
Welcome to PF, Luscinia! :smile:

No mistake, your answer to (d) is correct.
It is arcsin(x-1) + C and it evaluates to ∏/6.
(Sorry sandy.bridge.)
 
  • #4
798
1
Oops, you're right. I presume that your issue was with regards to your lower and upper limits. Did you remember to alter them during substitution?
 
  • #5
17
0
Thanks for the quick answer!
I'm not sure why I need to replace the upper and lower limit since the final integral for ∫dx/(√(2x-x2) uses the variable x as well.
(Though I honestly forgot about them and I'm not too sure if I know how to do it properly)
I had used (x-1)=sinu as substitution when I had integrated so arcsin(x-1)=arcsin(sinu)=u

(1/2-1)=sinu
arcsin(-1/2)=u=-∏/6 (lower limit)

(1-1)sinu
arcsin(0)=u=0 (upper limit)

using FTC:
0--∏/6=∏/6?

Maybe my teacher made a mistake?
 
  • #6
798
1
Okay, I just did the question and got the same answer as you. Your teacher must have given you a false answer.
 

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