# Change of variables to polar coordinates with constant $r$ evaluate definite integral.

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?

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

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?