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

• ashnicholls
In summary, the conversation discusses how to integrate the expression \int_a^4\sqrt{16-r^2}rdr using substitution and integration by parts. The correct answer is \frac{1}{3}(16-r^2)^{\frac{3}{2}} with limits a to 4. The conversation also briefly mentions converting the limits when making a change of variable.

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

which one though? u=16-r^2

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

Cheers

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

with limits a to 4

ashnicholls said:
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.

What so you just ignore the first r?

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

That is correct.

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

cheers

$$u = 16 - r^2$$

$$du = - 2rdr$$

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

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

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

ashnicholls said:
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.

ashnicholls said:
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?

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

and the lower limit is root(16 - u)

ashnicholls said:
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.

or yes course

so lower limit is 16 - a^2

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

Last edited:
That seems right.

are you sure?

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

ashnicholls said:
are you sure?

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