use polar coordinates to evaluate $ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $

Question

$$ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $$ converting to polar coordinates :

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2(cos^2 + sin^2)}} dydx $$ using trig substitution $cos^2 + sin^2 = 1$

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2}} dydx = \int_0^1 \int_0^{\sqrt{1-r^2}} e^r drdx = \int_0^1 e^{\sqrt{1-r^2} } - 1 \space drdx = e^1 - 1 - e^1 = -1 $$ Is this logic correct ?

Answer 1

In polar coordinates, $x=r\cos \theta$ and $y= r\sin \theta$. Before we compute anything, we need our Jacobian, which tells us the area element for our coordinate system. In this case, we will have $dA = r dr d\theta$. You can verify this by direct calculation.

Now, let's look at the limits. Notice that in $y$ we go from $0$ to $y=\sqrt{1-x^2}$ and then in $x$ we go from $0$ to $1$. So the region is a semi-circle in the top-right quadrant (the limits 'trace out' the blue section in the picture)

In polar coordinates, that is we go from $0$ to $1$ radially and $0$ to $\frac{\pi}{2}$ for our angle. So we have $$\int_0^1 \int_0^{\sqrt{1-x^2}}e^{\sqrt{x^2+y^2}} dydx = \int_0^1 \int_0^{\frac{\pi}{2}} e^r rd\theta dr$$ Notice there is no $\theta$ dependence in the integral so we can take out the $\frac{\pi}{2}$ to get $$\frac{\pi}{2} \times \int_0^1 re^r dr = \frac{\pi}{2}$$ That comes from integration by parts, which you can do yourself for verification.

Answer 2

• If you draw that region in the plane you'll see that its the part of the unit disk contained in the first quadrant i.e calling the domain $D$ then $D = \{(r, \theta): 0 \leq r \leq ?, 0 \leq \theta \leq ?\}$. Fill those in.
• Now for the integrand, recall that you just change $x,y$ to their polar form and tac on the $r$ from the jacobian which accounts for the relationship between areas when making the coordinate change i.e,

$$\textbf{Area}= \int_{\theta=?}^{\theta=?} \int_{r=?}^{r=?} e^r \cdot r \ dr \ d\theta$$

Answer 3

What you have is a mess.

you can't say $\cos^2 + \sin^2 = 1$ you need some object.

When you convert to polar coordinates, your coordinate system will be in terms of $r, \theta.\ \ \ dx\ dy$ will be replaced

Start with.

$x = r \cos\theta\\y=r\sin\theta$

your limits.

$y \le \sqrt {1-x^2}\\ r\sin\theta \le \sqrt {1-r\cos\theta}\\ r^2\sin^2 \theta + r^2\cos^2 \theta le 1$ r le 1$

$y\ge 0\\ r\sin\theta \ge 0\\ \sin\theta \ge 0\\ 0\le \theta \le \pi $

$x>0\\ \cos\theta \ge 0\\ 0\le \theta \le \frac {\pi}{2} $

The jacobian... I am not sure I am ready educate you on this. Look it up. $dy\ dx = r\ dr\ d\theta$

$\int_0^{\frac {\pi}{2}}\int_0^1 re^r \ dr\ d\theta$

$\frac {\pi}{2} (r-1) e^r|_0^1\\ \frac {\pi}{2}$

Answer 4

The crucial thing to realise is that the integration is over the quarter of the unit circle with both $x$ and $y$ positive. We are told to use polar coordinates $x= r \cos \theta$, $y=\sin \theta $ and $dxdy=r dr d \theta$ and so the integral becomes \begin{eqnarray*} \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx = \int_0^{\frac{ \pi}{2}} d \theta \int_0^1 r e^r dr =\frac{\pi}{2}. \end{eqnarray*}

Answer 5

$$ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $$

Notice that we are finding the region is a quadrant of a circle with radius 1

$$\int_0^{\pi/2} \int_0^{1} r \cdot e^{\sqrt{r^2}} drd\theta = \int_0^{\pi/2} \int_0^{1} r \cdot e^r dr d\theta$$

Integrating the inner integral by parts

$$\int r \cdot e^r dr = r \cdot e^r-e^r$$

And from $0$ to $1$, this is $1$

Now our integral is

$$\int_0^{\pi/2} 1 d\theta$$

Which clearly evaluates to $\pi/2$