Sketch the region whose area is given by the integral and evaluate the following integral: $\int_{\pi}^{2\pi} \int_{2}^{6} 4r \, dr \, d\theta$

Neetesh Kumar | November 28, 2024

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

Step 1: Understanding the Integral

The given integral is in polar coordinates, where the general form for a double integral is:

$\int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r, \theta) \, r \, dr \, d\theta$

In this case, the function is $4r$, and the limits for $r$ and $\theta$ are as follows:

• $r$ ranges from $2$ to $6$.
• $\theta$ ranges from $\pi$ to $2\pi$.

This indicates that the region of integration is a sector of a circle with:

• Radius between $2$ and $6$.
• Angle between $\pi$ and $2\pi$ (i.e., the bottom half of a circle).

Step 2: Sketching the Region

The region corresponds to a sector of an annular (ring-shaped) area. The two key characteristics of the region are:

• The inner radius is $r = 2$.
• The outer radius is $r = 6$.
• The angle sweeps from $\theta = \pi$ to $\theta = 2\pi$, which covers the bottom half of the circle (from left to right).

To sketch:

• Draw two concentric circles, one with radius $2$ and the other with radius $6$.
• Shade the region between these two circles, from angle $\pi$ to $2\pi$ (the lower half of the ring).

This gives you the exact region described by the limits of the integral.

Step 3: Evaluating the Integral

Now, let’s evaluate the integral:

$\int_{\pi}^{2\pi} \int_{2}^{6} 4r \, dr \, d\theta$

Inner Integral: Integration with respect to $r$

We first focus on the inner integral:

$\int_{2}^{6} 4r \, dr$

The integral of $4r$ is straightforward:

$\int 4r \, dr = 2r^2$

Now, evaluate this from $r = 2$ to $r = 6$:

$2r^2 \Big|_{2}^{6} = 2(6^2) - 2(2^2) = 2(36) - 2(4) = 72 - 8 = 64$

Outer Integral: Integration with respect to $\theta$

Next, we handle the outer integral with respect to $\theta$:

$\int_{\pi}^{2\pi} 64 \, d\theta$

Since $64$ is a constant, the integral simplifies to:

$64 \times (\theta \Big|_{\pi}^{2\pi}) = 64 \times (2\pi - \pi) = 64 \times \pi$

Thus, the value of the integral is:

$64\pi$

Final Answer:

The area of the region is given by the integral, and its evaluation yields: $\boxed{64\pi}$

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