To evaluate the iterated integral

$\int_0^5 \int_0^{\sqrt{25 - x^2}} e^{-(x^2 + y^2)} \, dy \, dx$

we can simplify this by converting to polar coordinates.

Step 1: Set Up the Integral in Polar Coordinates

In polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$. The area element $dx \, dy$ becomes $r \, dr \, d\theta$.

Also, the exponent $-(x^2 + y^2)$ becomes $-r^2$.

Step 2: Determine the Limits for $r$ and $\theta$

The region of integration is given by $0 \leq x \leq 5$ and $0 \leq y \leq \sqrt{25 - x^2}$. This describes the area of a quarter-circle in the first quadrant with radius 5.

Therefore:

• $r$ ranges from $0$ to $5$,
• $\theta$ ranges from $0$ to $\frac{\pi}{2}$ (first quadrant).

Step 3: Set Up the Integral in Polar Coordinates

Substitute the expressions and limits into the integral:

$\int_0^{\pi/2} \int_0^5 e^{-r^2} \cdot r \, dr \, d\theta$

Step 4: Evaluate the Integral

Separate the integrals over $r$ and $\theta$:

$\int_0^{\pi/2} d\theta \int_0^5 r e^{-r^2} \, dr$

1. Evaluate the $\theta$-integral: $\int_0^{\pi/2} d\theta = \frac{\pi}{2}$

2. Evaluate the $r$-integral: To integrate $\int_0^5 r e^{-r^2} \, dr$, use the substitution $u = r^2 \Rightarrow du = 2r \, dr \Rightarrow \frac{du}{2} = r \, dr$.

   Then the limits change as follows:

   • When $r = 0$, $u = 0$.
   • When $r = 5$, $u = 25$.

   So the $r$-integral becomes: $\int_0^5 r e^{-r^2} \, dr = \frac{1}{2} \int_0^{25} e^{-u} \, du$

   Now integrate: $\frac{1}{2} \int_0^{25} e^{-u} \, du = \frac{1}{2} \left[ -e^{-u} \right]_0^{25} = \frac{1}{2} \left( 1 - e^{-25} \right)$

Step 5: Combine the Results

Now, multiply the results of the $r$-integral and $\theta$-integral:

$\frac{\pi}{2} \cdot \frac{1}{2} \left( 1 - e^{-25} \right) = \frac{\pi}{4} \left( 1 - e^{-25} \right)$

Final Answer

$\int_0^5 \int_0^{\sqrt{25 - x^2}} e^{-(x^2 + y^2)} \, dy \, dx = \frac{\pi}{4} \left( 1 - e^{-25} \right)$

---

Would you like more details or have any questions?

Related Questions:

1. How does one generally approach changing to polar coordinates in integrals?
2. What is the significance of the exponential function in integrals over radial distances?
3. Why does the integral for $\theta$ go from $0$ to $\pi/2$ in this case?
4. How would the limits change if the region covered a different quadrant?
5. What other methods could simplify integrals with circular symmetry?

Tip:

Always check the region of integration carefully before switching to polar coordinates to ensure that the limits for $r$ and $\theta$ are set correctly.