Let's walk through the solution for evaluating the integral over the given region $R$ by using polar coordinates.

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1. Understand the region $R$:

The region of integration $R$ is defined as:
$R = \left\{ (x, y) : x^2 + y^2 \leq 1, \quad x \geq 0, \quad y \geq 0 \right\}$

This describes the quarter-circle in the first quadrant with radius $1$. That is, it includes points $(x, y)$ such that the distance from the origin is at most 1, but only in the first quadrant.

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2. Set up the integral in polar coordinates:

We convert the integral to polar coordinates, using: $x = r \cos \theta, \quad y = r \sin \theta, \quad dA = r \, dr \, d\theta$ For a quarter-circle in the first quadrant:

• $r$ ranges from $0$ to $1$.
• $\theta$ ranges from $0$ to $\frac{\pi}{2}$.

Now, we rewrite the integrand.
The integrand is: $\frac{1}{\sqrt{36 - x^2 - y^2}}$ In polar coordinates: $x^2 + y^2 = r^2 \quad \Rightarrow \quad 36 - x^2 - y^2 = 36 - r^2$ Thus, the integrand becomes: $\frac{1}{\sqrt{36 - r^2}}$

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3. Rewrite the integral:

The original integral over $R$ becomes: $\int_0^{\pi/2} \int_0^1 \frac{1}{\sqrt{36 - r^2}} \, r \, dr \, d\theta$

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4. Evaluate the integral:

Step 1: Integrate with respect to $r$.

We need to compute: $\int_0^1 \frac{r}{\sqrt{36 - r^2}} \, dr$ Use the substitution:
$u = 36 - r^2 \quad \Rightarrow \quad du = -2r \, dr$.
Thus: $\int_0^1 \frac{r}{\sqrt{36 - r^2}} \, dr = -\frac{1}{2} \int_{36}^{35} \frac{-du}{\sqrt{u}} = \frac{1}{2} \int_{35}^{36} u^{-1/2} \, du$ Evaluating the integral: [ \frac{1}{2} \left[ 2u^{1/2} \right]{35}^{36} = \left[ \sqrt{u} \right]{35}^{36} = \sqrt{36} - \sqrt{35} = 6 - \sqrt{35} ]

Step 2: Integrate with respect to $\theta$.

Now we integrate with respect to $\theta$: $\int_0^{\pi/2} (6 - \sqrt{35}) \, d\theta = (6 - \sqrt{35}) \cdot \left[ \theta \right]_0^{\pi/2} = (6 - \sqrt{35}) \cdot \frac{\pi}{2}$

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5. Final Answer:

The value of the integral is: $\frac{\pi}{2} (6 - \sqrt{35})$

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Summary:

The integral evaluates to: $\frac{\pi}{2} (6 - \sqrt{35})$

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Would you like me to provide more detailed steps or further clarifications?

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Related Questions:

1. How do you convert an integral from Cartesian to polar coordinates?
2. What is the geometric meaning of the limits in polar coordinates?
3. How does the Jacobian factor $r$ appear in polar coordinates?
4. How do you integrate functions involving $\sqrt{a^2 - r^2}$?
5. What happens if the region of integration covers a full circle instead of a quadrant?

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Tip:

When switching to polar coordinates, remember that the area element changes from $dx \, dy$ to $r \, dr \, d\theta$, which accounts for the stretching near the origin.