# Double Integral

Evaluate the integral by changing to polar coordinates

$$\int\int arctan(y/x)$$

Given that

0 $$\leq$$ x $$\leq$$ 1 and 0 $$\leq$$ y $$\leq$$ x

Now I've changed the integral to

$$\int\int \theta r dr d\theta$$

Such that 0 $$\leq$$ $$\theta$$ $$\leq$$ $$\frac{\pi}{2}$$ and 0 $$\leq$$ r $$\leq$$ $$\sqrt{2}$$

And evaluating this I get $$\frac{\pi^{2}}{8}$$

I don't think this is correct though, I have found out that the answer is $$\frac{\pi}{8}$$ - 0.25ln(2)

Can someone show me where I'm going wrong please, thanks!

Sorry for the poor Latex-ing I've never used it before.

tiny-tim
Homework Helper
Hi Juggler123!

(have an integral: ∫ and a pi: π and a theta: θ and a ≤ )

Sorry, but your limits are toooootally wrong.

0 ≤ x ≤ 1 and 0 ≤ y ≤ x is a 45º triangle with base 1.

So your limits are 0 ≤ θ ≤ 45º and 0 ≤ r ≤ … (depends on θ ) ?

Would it be correct to say that the upper limit for r is 1/cos$$\theta$$ ?

tiny-tim
Homework Helper
Hi Juggler123!

(what happened to that θ i gave you? )
Would it be correct to say that the upper limit for r is 1/cos$$\theta$$ ?

Noooo.

How can I go about finding the upper limit then? I used the fact that the maximum x=1=max.y then said x=rcos(theta) and rearranged, where have I gone wrong??

Mark44
Mentor
Hint: In your triangular region, every x value on the right boundary has the same value.

Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?

tiny-tim
Homework Helper
oops!

How can I go about finding the upper limit then? I used the fact that the maximum x=1=max.y then said x=rcos(theta) and rearranged, where have I gone wrong??

oh no, I had my diagram the wrong way up

you were right, it is 0 ≤ r ≤ 1/cosθ.

Sorry.

ok, now integrate ∫∫ θr drdθ between those limits.

Mark44
Mentor
Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?
Because, on that boundary x = 1 ==> rcos(θ) = 1, so r = 1/cos(θ).

It looks like you already understand this, based on another reply you gave.

Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x

Mark44
Mentor
Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x
This is actually pretty simple.
$$\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx$$
Edit: added missing x factor in right integral.
Hint: There is a trig function whose derivative wrt x is sec^2(x).

Last edited:
Isn't x/2cos(x)^2 equal to 0.5xsec(x)^2?? Not 0.5sec(x)^2

Mark44
Mentor
Right you are. The integral should be
$$\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx$$
This looks like a natural for integration by parts, with u = x and dv = sec^2(x)dx.

Yes! That works brilliant, thanks for all the help!