Evaluate the given integral by changing to polar coordinates: $\iint_R \sqrt{49 - x^2 - y^2} \, dA$ where $R = \{(x, y) | x^2 + y^2 \leq 49, x \geq 0 \}$.

Neetesh Kumar | November 28, 2024

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This is the solution to Math 1D
Assignment: 15.3 Question Number 3
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Step-by-step solution:

Step 1: Understanding the Region $R$

The region $R$ represents the right half of the disk $x^2 + y^2 \leq 49$ (a circle of radius 7). This means that the region is bounded by:

• The disk $x^2 + y^2 \leq 49$ with a radius of 7.
• The condition $x \geq 0$ restricts the region to the right half of the circle.

Thus, in polar coordinates:

• The radius $r$ ranges from 0 to 7, since the radius of the circle is 7.
• The angle $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ because we are covering the right half of the circle (the positive x-axis side).

Step 2: Converting to Polar Coordinates

In polar coordinates, we use the following transformations:

• $x = r \cos \theta$
• $y = r \sin \theta$
• The differential area element is $dA = r \, dr \, d\theta$

Now, substitute these into the integrand $\sqrt{49 - x^2 - y^2}$. Using $x^2 + y^2 = r^2$, we have:

$\sqrt{49 - x^2 - y^2} = \sqrt{49 - r^2}$

Thus, the integral becomes:

$\iint_R \sqrt{49 - x^2 - y^2} \, dA = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^7 \sqrt{49 - r^2} \cdot r \, dr \, d\theta$

Step 3: Evaluating the Inner Integral

We need to evaluate the inner integral with respect to $r$:

$\int_0^7 r \sqrt{49 - r^2} \, dr$

To solve this, we use the substitution method. Let:

$u = 49 - r^2$

Then, $du = -2r \, dr$, which gives $r \, dr = -\frac{1}{2} \, du$. The limits of integration change as follows:

• When $r = 0$, $u = 49$.
• When $r = 7$, $u = 0$.

Substituting into the integral, we get:

$\int_0^7 r \sqrt{49 - r^2} \, dr = \int_{49}^0 \sqrt{u} \left( -\frac{1}{2} \, du \right)$

This simplifies to:

$= \frac{1}{2} \int_0^{49} \sqrt{u} \, du$

We know that $\sqrt{u} = u^{1/2}$, so the integral of $u^{1/2}$ is:

$\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

Now evaluate from 0 to 49:

$\frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^{49} = \frac{1}{3} \left[ u^{3/2} \right]_0^{49}$

Substitute the limits:

$= \frac{1}{3} \left[ 49^{3/2} - 0^{3/2} \right] = \frac{1}{3} \cdot 49^{3/2}$

Since $49^{3/2} = (7^2)^{3/2} = 7^3 = 343$, we have:

$= \frac{343}{3}$

Thus, the value of the inner integral is:

$\frac{343}{3}$

Step 4: Evaluating the Outer Integral

Now, evaluate the outer integral with respect to $\theta$:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta = \theta \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi$

Step 5: Combine the results:

Finally, multiply the result of the inner integral by the result of the outer integral:

$\left( \frac{343}{3} \right) \times \pi = \frac{343\pi}{3}$

Final Answer:

The value of the given integral is: $\boxed{\frac{343\pi}{3}}$

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