Trig Substitution and Volume of Solids

• Luscinia
In summary: Thanks for the help!In summary, the definite integral of ∫dx/(√(2x-x2)) is arcsin(x-1) + C and evaluates to ∏/6.
Luscinia
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)

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!

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

Welcome to PF, Luscinia!

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.

1. What is trig substitution?

Trig substitution is a technique used in calculus to simplify integrals involving expressions with trigonometric functions. It involves replacing a variable in the integral with a trigonometric function to make the integral easier to solve.

2. When should I use trig substitution?

Trig substitution is typically used when the integral involves expressions with square roots and/or quadratic terms. It is also used when the integrand contains a combination of trigonometric and algebraic functions.

3. How do I choose the appropriate trig substitution?

The choice of trig substitution depends on the form of the integrand. For integrands involving a2 - x2, use the substitution x = a sin(θ). For integrands involving a2 + x2, use the substitution x = a tan(θ). And for integrands involving x2 - a2, use the substitution x = a sec(θ).

4. What is the volume of a solid generated by revolving a region using trig substitution?

The volume of a solid generated by revolving a region using trig substitution can be found using the formula V = π∫ba(f(x))2dx. This formula can be derived using the disk method or the shell method, depending on the shape of the region being revolved.

5. Can trig substitution be used for all types of integrals?

No, trig substitution is only applicable for integrals involving trigonometric functions. For other types of integrals, other integration techniques such as integration by parts or partial fractions may be more appropriate.

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