Tricky Cartesian to Polar Change of Variables Integral

[JasonJo]

Hmm, I can't seem to get this double integral transformation:

int(limits of integration are 0 to 3) int (limits of int are 0 to x) of (dy dx)/(x^2 + y^2)^(1/2)

and i need to switch it to polar coordinates and then evaluate the polar double integral.

i sketched the region over which i am integrating and it isn't a circular region (which is why I guess this was assigned).

but i can't the limits of integration for the theta angle. the radius goes from 0 to 3, but i can't get the theta angle limits of integration.

the answer is 3ln(sqrt(2) + 1))
i can see where the 3 comes from, but i don't know where i would get the ln term.

thank you in advance

 

[Galileo]

Have you sketched the region in the xy-plane? If you do you'll find that the radius depends on the angle (since it's not a circular region like you said). For example, r goes from 0 to 3 on the x-axis (theta=0), but it goes from 0 to sqrt(18) for if theta=Pi/4.

 

[JasonJo]

yeah , i had that the radius goes from 0 to sqrt(18) and that theta goes from 0 to 45*
however, after you do the change of variables, (ie, dydx = r dr d(theta))
im double integrating 1, and i get 3*sqrt(2)/2, but the real answer is 3 ln(sqrt(2) + 1))
any other hints or clues?

 

[Galileo]

If you let r go from 0 to sqrt(18) regardless of theta, you will get a section of a disk. r is dependent upon theta.

You are right to let theta vary from 0 to Pi/4. But given an angle theta, between which values should r vary? Hint: look at picture.

 

[HallsofIvy]

The horizontal line y= 3 is $r sin(\theta)= 3$ or $r= \frac{3}{cos(\theta)}$. The integral is
$$\int_{\theta= 0}^{\frac{\pi}{2}} \int_{r= 0}^{\frac{3}{cos(\theta)}}drd\theta= 3\int_{\theta= 0}^{\frac{\pi}{2}}\frac{d\theta}{cos(\theta)}$$

To do that integral, multiply numerator and denominator by $cos(\theta)$ and let $u= cos(\theta)$.

 

[JasonJo]

Doh!

Thanks Guys!