Converting an integral from polar to cartesian Question concerning definite integrals. Lets say we have some integral in cartesian coordinates (like the integral of 

$$ \int^{3/4}_0 \sqrt{1+y^2} \space dy$$ 

I completely understand how to convert the function itself into polar coordinates as well as how to find the indefinite integral of a polar equation. What I do not understand is how to reevaluate the limits of integration. Any help would be appreciated.
Normally an integral in Cartesian coordinates is a two dimensional integral over some region of the plane, not a one dimensional interval as in your example.  If it is a simple rectangular region $\int_a^b\int_c^d f(x,y)dx\ dy$  it is hard to express in polar coordinates.  That is a strong reason to stay in Cartesian coordinates.  If the region is simple in polar coordinates, like a circle at the origin $\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} f(x,y) dx \ dy$ it becomes simple in polar coordinates $\int_0^{2 \pi}\int _0^R f(r,\theta)r\; dr\; d\theta$ where I have abused notation by using $f$ for both versions, irrespective of the arguments.  The second is much simpler for circular regions, so is a candidate when the region is simple in polar coordinates.
@Louie I would say, first of all you have to make it double integral in order to use substitution using polar coordinate. For your case you can assume

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$ 
or it can be
$$ I = \int^{3/4}_0 \sqrt{1+x^2} \space dy$$

both are same as only x variable is replaced by y. Now, multiply both of the above equations, you would get 
$$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2} \sqrt{1+y^2} \space dxdy$$
or 
$$ I^2 = \int^{3/4}_0 \int^{3/4}_0 \sqrt{1+x^2+y^2+x^2y^2}  \space dxdy$$

Now use the polar coordinate substitution as

$x^2+y^2=r^2$

and the limit of one integral will be for $\theta$ from $0$ to $2\pi$ and for the other integral it will be from $0$ to $r$.

After evaluation, take sqaure root of $I^2$ to end up with your integral solution which is for

$$ I = \int^{3/4}_0 \sqrt{1+y^2} \space dy$$  

I hope it would help you.