# Homework Help: Trig Substitution and Volume of Solids

1. Nov 9, 2011

### Luscinia

2 Questions here! (I'm not exactly sure if it's allowed, but I want to avoid posting too many threads)

1. The problem statement, all variables and given/known data
1)Find the following definite integrals by using a trigonometric substitution:
d)1/21dx/(√(2x-x2)

2. Relevant equations
x=asinu

3. 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.

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Shell:V= 2∏abxhdx
Disc: V=∏abr22-r12dx

3. 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!

2. Nov 9, 2011

### sandy.bridge

For d) It is not
$$arcsin(x-1)$$

3. Nov 9, 2011

### I like Serena

Welcome to PF, Luscinia!

It is arcsin(x-1) + C and it evaluates to ∏/6.
(Sorry sandy.bridge.)

4. Nov 9, 2011

### 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?

5. Nov 9, 2011

### Luscinia

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. Nov 9, 2011

### sandy.bridge

Okay, I just did the question and got the same answer as you. Your teacher must have given you a false answer.