# How to integrate sqrt(16-r^2) from a to 4

1. Mar 10, 2007

### ashnicholls

$$\int_a^4\sqrt{16-r^2}rdr$$

How do you integrate this?

Is it by parts, and then by substitution for the root bit?

Cheers Ash

Last edited: Mar 10, 2007
2. Mar 10, 2007

### neutrino

If $\int_{a}^{4}{r\sqrt{16-r^2}}dr$ is your integral, a substitution would suffice.

3. Mar 10, 2007

### ashnicholls

which one though? u=16-r^2

and then work r out in terms u, and then integrate by parts?

Cheers

4. Mar 10, 2007

### ashnicholls

$$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$

with limits a to 4

5. Mar 10, 2007

### neutrino

Yes

In a way, yes. What you actually do is differentiate the expression u = 16 - r2

There is no need for that...follow the steps mentioned above and you'll know why.

6. Mar 10, 2007

### ashnicholls

What so you just ignore the first r?

7. Mar 10, 2007

### neutrino

That is correct.

8. Mar 10, 2007

### ashnicholls

yes just worked that out on a website, but what happens to the first r then?

cheers

9. Mar 10, 2007

### neutrino

$$u = 16 - r^2$$

$$du = - 2rdr$$

Can you spot an rdr somewhere in the original integral? Also, remember, when you're making a change of variable, the limits change accordingly.

10. Mar 10, 2007

### ashnicholls

sorry isnt it minus $$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$

11. Mar 10, 2007

### ashnicholls

yes done it now thanks, but how do you convert the limits?

12. Mar 10, 2007

### neutrino

Yes, that's right. Sorry I missed that.

It can also be +, but then, the limits have to interchanged.

Well, you know how u and r are related. What values would u take when r has those values?

13. Mar 10, 2007

### ashnicholls

oh so the upper limit is very u = 16 - 4^2 = 0

and the lower limit is root(16 - u)

14. Mar 10, 2007

### neutrino

Yes

Why would it be that. Put r=a.

15. Mar 10, 2007

### ashnicholls

or yes course

so lower limit is 16 - a^2

16. Mar 10, 2007

### ashnicholls

So is this right if someone doesnt mind checking.

$$\int_a^4\sqrt{16-r^2}rdr$$

using $$u = 16 - r^2$$ which converts to

$$du = - 2rdr$$

so -du/2r = dr

so integral becomes

the integral of - (root u)/2 with lower limit 16-a^2 and upper limit 0

which converts to

the integral of (root u)/2 with lower limit 0 and upper limit 16-a^2

which becomes [((u^(3/2))/3] with lower limit 0 and upper limit 16-a^2

which becomes ((16-a^2)^3/2)/3

Is that right, I hope you all understand it,

I dont understand that tex thing very well.

cheers Ash

Last edited: Mar 10, 2007
17. Mar 10, 2007

### neutrino

That seems right.

18. Mar 10, 2007

### ashnicholls

are you sure?????

19. Mar 10, 2007

put r=4 sinx,then dr= 4 cosx dx,16-r2=16(1-sin{square}x).
proceed.Work it out & tell me

20. Mar 10, 2007

### HallsofIvy

Staff Emeritus
Yes, he's sure! Aren't you?

That leads to a more difficult integral Did you notice the "r" outside the square root? The way he did it to begin with was completely correct.