Trig Substitution and Volume of Solids

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Homework Help Overview

The discussion revolves around finding definite integrals using trigonometric substitution and calculating volumes of solids generated by revolving regions bounded by curves. The subject area includes calculus concepts related to integration and volume calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the evaluation of a definite integral involving a trigonometric substitution and the calculation of volumes using different methods. There is uncertainty about the correctness of limits of integration and the application of the Fundamental Theorem of Calculus.

Discussion Status

Some participants have provided guidance regarding the evaluation of the integral and the importance of adjusting limits during substitution. There is a mix of interpretations regarding the correctness of the original poster's answer compared to the teacher's response.

Contextual Notes

Participants note the potential for confusion regarding the limits of integration and the implications of using substitutions in definite integrals. There is also mention of discrepancies between calculated answers and those provided by the teacher.

Luscinia
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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!
 
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For d) It is not
arcsin(x-1)
 
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.)
 
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?
 
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?
 
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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