Use the change of variables $x = \cos\theta$ and $y=\sin\theta$ to find $$\int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} \sqrt{x^2 + y^2} \,dy\,dx$$

Attempt 1: Assuming coordinates are supposed to be $\,dr\,d\theta$.

As $r=1$ (constant), the Jacobian determinant turns the integral to $$\iint0\,dr\,d\theta$$ Is this the way to go?

Attempt 2: Differentiating x and y with respect to $\theta$, $$dx=-\sin\theta d\theta$$ $$dy=\cos\theta d\theta$$

But when trying to evaluate limits for $x$ we get $$2=\cos\theta$$ But this can't be possible??

Also does an integral increment of $d\theta^2$ make sense?

  • $\begingroup$ Nope, $r$ is not constant. $\endgroup$
    – user65203
    Nov 17, 2020 at 8:26

2 Answers 2


I think you'd better again read your textbook with more effort, you apparently misled the concept of changing variables.

The integration domain is $$ \bigg\{{(x,\,y) | 0\leq x\leq 2,\;0\leq y\leq \sqrt{2x-x^2}}\bigg\}, $$ which is the upper semi-disc enclosed by the circle $(x-1)^2 + y^2 = 1$.

By letting $$ \begin{cases} x = r\cos\theta \\ y=r\sin\theta, \end{cases} $$ since $0\leq x\leq 2$ and $y\geq 0$, it follows that $0\leq\theta\leq\frac{\pi}{2}$, also, since $y\leq\sqrt{2x-x^2}$, it follows that $r\leq 2\cos\theta$.

Therefore, by changing the variables, the original integral becomes $$ \int_{0}^{\frac{\pi}{2}}\int_{0}^{2\cos\theta}r^2\,dr\,d\theta. $$

Note: In your way of changing variables, you should note that $r$ is not a constant. You are not integrating over a circle but a semi-disc.


Please check the limit of integration to understand the region you are integrating over.

$0 \leq y \leq \sqrt{2x-x^2}$

$y = \sqrt{2x-x^2} \implies y^2 = 2x - x^2 \implies (x-1)^2+y^2 = 1$

So the region is a circle with center at $(1,0)$ and radius $1 \, (0 \leq x \leq 2)$ .

In polar coordinates, this circle is represented as $2 \cos \theta$.

So the limits of integration will be $0 \leq r \leq 2 \cos \theta$ and $0 \leq \theta \leq \pi/2$ (as $y \geq 0)$.

In polar coordinates, $x = r \cos \theta, y = r \sin \theta$. Substitute in your integrand.

Can you take it from here?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.