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Question:

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?

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

2 Answers 2

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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.

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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?

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