Solve $\int_0^1 \int_{y}^{\sqrt y}\sqrt{x^2+y^2} \,dx \,dy$ using polar coordinates.

Question

Q: Solve $\int_0^1 \int_{y}^{\sqrt y}\sqrt{x^2+y^2} \,dx \,dy$ using polar coordinates.

I'm starting to learn this. I tried the following:

By drawing the graphs, I thought that I can say: $\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}$. Then if i fix an angle $\theta$, the radius $r$ will varie from $0$ to the distance between the origin and the intersection of the line that goes from the origin with angle $\theta$ ($x = tan(\theta)*y) $ and the curve $\sqrt y = x$.

The intersection occurs when $x = tan(\theta)*y = \sqrt y$, that is, $y = \dfrac{1}{tan^2(\theta)}$, and $x = \dfrac{1}{tan(\theta)}$. So, the radius is:

$r = \sqrt{x^2+y^2} = \sqrt{ \dfrac{1}{tan^2(\theta)} + \dfrac{1}{tan(\theta)}} = \sqrt{ \dfrac{tan (\theta) + 1}{tan^2(\theta)}}$.

Ok, so $0 < r < \sqrt{ \dfrac{tan (\theta) + 1}{tan^2(\theta)}}$.

And then we have that: $$\int_0^1 \int_{y}^{\sqrt y}\sqrt{x^2+y^2} \,dx \,dy = \int_{\pi/4}^{\pi/2} \int_0^{\sqrt{ \dfrac{tan (\theta) + 1}{tan^2(\theta)}}} r^2 \, dr \, d\theta$$.

But i think that this is a wrong (or very difficult) way to solve this. Can you say what i missed, or other ways to solve this? Thanks!

Answer 1

You have an algebra error here:

$$r = \sqrt{x^2+y^2} = \sqrt{ \dfrac{1}{\tan^2(\theta)} + \dfrac{1}{\tan(\theta)}} = \sqrt{ \dfrac{\tan (\theta) + 1}{\tan^2(\theta)}}$$

It should be $$r = \sqrt{x^2+y^2} = \sqrt{ \dfrac{1}{\tan^2(\theta)} + \dfrac{1}{\tan^4(\theta)}} = \frac{\sqrt{ \tan^2\theta + 1}}{\tan^2\theta} = \frac{\cos\theta}{\sin^2\theta}$$ The integral isn't too hard to do from there.

Hint:

Substitute $u = 1/\sin(\theta)$.

Answer 2

So you forgot to square when you substituted in $x$ and $y$.
Remember your trig identities: $$\begin{align}\sin^2\theta + \cos^2\theta =& 1\\ \frac{1}{\cos^2\theta}\cdot(\sin^2\theta + \cos^2\theta) =& \frac{1}{\cos^2\theta}\cdot1\\ \tan^2\theta + 1=& \frac{1}{\cos^2\theta} = \sec^2\theta\end{align}$$

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So, $$\begin{align*} &\sqrt{\frac{1}{\tan^4\theta} + \frac{1}{\tan^2\theta}}\hspace{10cm}\\ =&\sqrt{\frac{1 + \tan^2\theta}{\tan^4\theta}}\\ =&\frac{\sqrt{1 + \tan^2\theta}}{\sqrt{\tan^4\theta}}\\ =&\frac{\sqrt{\sec^2\theta}}{\sqrt{\tan^4\theta}}\\ =&\frac{\sec\theta}{\tan^2\theta}\\ =&\frac{\cos\theta}{{\sin}^2\theta}\end{align*}$$ From there it should be more doable