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

  • #1
[tex]\int_a^4\sqrt{16-r^2}rdr[/tex]

How do you integrate this?

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

Cheers Ash
 
Last edited:

Answers and Replies

  • #2
2,063
2
If [itex]\int_{a}^{4}{r\sqrt{16-r^2}}dr[/itex] is your integral, a substitution would suffice.
 
  • #3
which one though? u=16-r^2

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

Cheers
 
  • #4
Is the answer:
[tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]

with limits a to 4
 
  • #5
2,063
2
u=16-r^2
Yes

and then work r out in terms u,
In a way, yes. What you actually do is differentiate the expression u = 16 - r2

and then integrate by parts?
There is no need for that...follow the steps mentioned above and you'll know why.
 
  • #6
What so you just ignore the first r?
 
  • #7
2,063
2
Is the answer:
[tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]
with limits a to 4
That is correct.
 
  • #8
yes just worked that out on a website, but what happens to the first r then?

cheers
 
  • #9
2,063
2
[tex]u = 16 - r^2[/tex]

[tex]du = - 2rdr[/tex]

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
sorry isnt it minus [tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]
 
  • #11
yes done it now thanks, but how do you convert the limits?
 
  • #12
2,063
2
sorry isnt it minus [tex]\frac{1}{3}(16-r^2)^{\frac{3}{2}}[/tex]
Yes, that's right. Sorry I missed that.

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

yes done it now thanks, but how do you convert the limits?
Well, you know how u and r are related. What values would u take when r has those values?
 
  • #13
oh so the upper limit is very u = 16 - 4^2 = 0

and the lower limit is root(16 - u)
 
  • #14
2,063
2
oh so the upper limit is very u = 16 - 4^2 = 0
Yes

and the lower limit is root(16 - u)
Why would it be that. Put r=a.
 
  • #15
or yes course

so lower limit is 16 - a^2
 
  • #16
So is this right if someone doesnt mind checking.

[tex]\int_a^4\sqrt{16-r^2}rdr[/tex]

using [tex]u = 16 - r^2[/tex] which converts to

[tex]du = - 2rdr[/tex]

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
 
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  • #17
2,063
2
That seems right.
 
  • #18
are you sure?????
 
  • #19
3
0
put r=4 sinx,then dr= 4 cosx dx,16-r2=16(1-sin{square}x).
proceed.Work it out & tell me
 
  • #20
HallsofIvy
Science Advisor
Homework Helper
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are you sure?????
Yes, he's sure! Aren't you? :approve:

put r=4 sinx,then dr= 4 cosx dx,16-r2=16(1-sin{square}x).
proceed.Work it out & tell me
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.
 
  • #21
3
0
sorry.he is right
 

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