Find the area of the region(s) between the given curves below on the given interval.

Question

y = 8 cos x,  y = 8 − 8 cos x from x = 0 to x = 𝜋

.

Answer 1

Let's solve the problem step by step to find the correct area between the curves:

1. **Find the points of intersection:**

Given equations:

\[ 8 \cos x = 8 - 8 \cos x \]

Solving for \(\cos x\):

\[

16 \cos x = 8

\]

\[

\cos x = \frac{1}{2}

\]

The solutions for \(\cos x = \frac{1}{2}\) within one period of \(2\pi\) are:

\[

x = \frac{\pi}{3} \text{ and } x = \frac{5\pi}{3}

\]

2. **Set up the integrals:**

- For \(0 \leq x \leq \frac{\pi}{3}\), the curve \(y = 8 \cos x\) is above \(y = 8 - 8 \cos x\).

- For \(\frac{\pi}{3} \leq x \leq \pi\), the curve \(y = 8 - 8 \cos x\) is above \(y = 8 \cos x\).

Therefore, the area \(A\) between the curves is:

\[

A = \int_{0}^{\frac{\pi}{3}} \left[(8 \cos x) - (8 - 8 \cos x)\right] \, dx + \int_{\frac{\pi}{3}}^{\pi} \left[(8 - 8 \cos x) - (8 \cos x)\right] \, dx

\]

Simplify the integrands:

\[

8 \cos x - (8 - 8 \cos x) = 16 \cos x - 8

\]

\[

(8 - 8 \cos x) - 8 \cos x = 8 - 16 \cos x

\]

So:

\[

A = \int_{0}^{\frac{\pi}{3}} (16 \cos x - 8) \, dx + \int_{\frac{\pi}{3}}^{\pi} (8 - 16 \cos x) \, dx

\]

3. **Evaluate the integrals:**

\[

\int_{0}^{\frac{\pi}{3}} (16 \cos x - 8) \, dx

\]

\[

\int_{0}^{\frac{\pi}{3}} 16 \cos x \, dx - \int_{0}^{\frac{\pi}{3}} 8 \, dx

\]

\[

16 \sin x \bigg|_{0}^{\frac{\pi}{3}} - 8 x \bigg|_{0}^{\frac{\pi}{3}}

\]

\[

16 \left(\sin \frac{\pi}{3} - \sin 0\right) - 8 \left(\frac{\pi}{3} - 0\right)

\]

\[

16 \left(\frac{\sqrt{3}}{2}\right) - \frac{8\pi}{3}

\]

\[

8\sqrt{3} - \frac{8\pi}{3}

\]

Now for the second integral:

\[

\int_{\frac{\pi}{3}}^{\pi} (8 - 16 \cos x) \, dx

\]

\[

\int_{\frac{\pi}{3}}^{\pi} 8 \, dx - \int_{\frac{\pi}{3}}^{\pi} 16 \cos x \, dx

\]

\[

8 x \bigg|_{\frac{\pi}{3}}^{\pi} - 16 \sin x \bigg|_{\frac{\pi}{3}}^{\pi}

\]

\[

8 (\pi - \frac{\pi}{3}) - 16 (\sin \pi - \sin \frac{\pi}{3})

\]

\[

8 \left(\frac{2\pi}{3}\right) - 16 \left(0 - \frac{\sqrt{3}}{2}\right)

\]

\[

\frac{16\pi}{3} + 8\sqrt{3}

\]

4. **Combine the results:**

Adding the two areas:

\[

A = \left(8 \sqrt{3} - \frac{8 \pi}{3}\right) + \left(\frac{16 \pi}{3} + 8 \sqrt{3}\right)

\]

\[A = 16 \sqrt{3} + \frac{8 \pi}{3} \]

Therefore, the correct area between the curves is:

\[

A = 16 \sqrt{3} + \frac{8 \pi}{3}

\]

Answer 2

8cosx = 8 - 8cosx; cosx = 1/2; x = π/3;

Area A = ∫₀^π/3((16cosx - 8)dx + ∫_π/3^π(8 - 16cosx)dx = (16sinx - 8x)₀^π/3 + (8x - 16sinx)_π/3^π = 8√3 - 8π/3 + 8π -

8π/3 + 8√3 = 16√3 + 8π/3