Evaluating an Integral in Polar Coordinates

PsychonautQQ
Messages
781
Reaction score
10

Homework Statement


Evalutate the double integral sin(x^2+y^2)dA between the region 1≥x^2+y^2≥49



The Attempt at a Solution


so r^2 = x^2 + y^2
dA = rdrdθ

so I can turn this into
double integral sin(r^2)rdrdθ
where the inner integral integrated with respect to dr goes from 1 to 7?
and then what are the parameters for integrating with respect to θ? 0 to 2pi?
Does this look correct?
 
Physics news on Phys.org
To clarify your question: it gives no boundary conditions for your \theta axis, so you are integrating over a disk of radius 7 (starting at 1). So yes your integral seems correct.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top