Question

Determine a definite integral that represents the area of the region enclosed by r = 8.

ᵇ∫₀ ______ dθ

Bounds: a = 0 to b = _____

Answer 1

The definite integral representing the area enclosed by the polar curve r = 8, with bounds from
`\(\theta = 0\)` to
`\(\theta = 2\pi\)`, is
`\((1)/(2) \int_(0)^(2\pi) 64 \, d\theta\)`, yielding an area of
`\(64\pi\)`.

To find the definite integral that represents the area of the region enclosed by the polar curve r = 8, you need to integrate
`\((1)/(2) r^2\)` with respect to
`\(\theta\)` over the given bounds. The formula for the area of a polar region is
`\((1)/(2) \int_(a)^(b) r^2 \, d\theta\)`.

For the curve r = 8, the definite integral would be:

`\[A = (1)/(2) \int_(0)^(b) (8)^2 \, d\theta\]`

You need to determine the upper limit b for the bounds. If you want to find the area of the entire region enclosed by the curve, you need to find where the curve completes one full revolution, which is
`\(2\pi\)` radians. So,
`\(b = 2\pi\)`.

The definite integral becomes:

`\[A = (1)/(2) \int_(0)^(2\pi) (8)^2 \, d\theta\]`

Simplify and solve this integral to find the area of the region enclosed by the polar curve r = 8.

To complete the expression, we can simplify the integral:

`\[ A = (1)/(2) \int_(0)^(2\pi) 64 \, d\theta \]`

Now, integrate with respect to
`\(\theta\)`:

`\[ A = (1)/(2) \left[ 64\theta \right]_(0)^(2\pi) \]`

Evaluate the expression at the upper and lower bounds:

`\[ A = (1)/(2) \left[ 64(2\pi) - 64(0) \right] \]`

Simplify:

`\[ A = (1)/(2) \cdot 128\pi \]\[ A = 64\pi \]`

So, the definite integral that represents the area of the region enclosed by the polar curve r = 8 with bounds from
`\(\theta = 0\)` to
`\(\theta = 2\pi\)` is
`\(64\pi\)`.