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

[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

 

[neutrino]

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

 

[ashnicholls]

which one though? u=16-r^2

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

Cheers

 

[ashnicholls]

Is the answer:
$$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$

with limits a to 4

 

[neutrino]

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.

 

[ashnicholls]

What so you just ignore the first r?

 

[neutrino]

Is the answer:
$$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$
with limits a to 4

That is correct.

 

[ashnicholls]

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

cheers

 

[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.

 

[ashnicholls]

sorry isn't it minus $$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$

 

[ashnicholls]

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

 

[neutrino]

sorry isn't it minus $$\frac{1}{3}(16-r^2)^{\frac{3}{2}}$$

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?

 

[ashnicholls]

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

and the lower limit is root(16 - u)

 

[neutrino]

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.

 

[ashnicholls]

or yes course

so lower limit is 16 - a^2

 

[ashnicholls]

So is this right if someone doesn't 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 don't understand that tex thing very well.

cheers Ash

 

[neutrino]

That seems right.

 

[ashnicholls]

are you sure?

 

[Padmashri]

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

 

[HallsofIvy]

are you sure?

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

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.

 

[Padmashri]

sorry.he is right