Area of Cardioid r=7(1+cosθ) Solved

Image Analysis

The image displays a polar equation $r = 7(1 + \cos(\theta))$ and four candidate graphs. The curve is a cardioid, a heart-shaped polar graph. Key features of $r = a(1 + \cos(\theta))$ include a cusp at the origin, a maximum reach of $2a$ along the positive x-axis (since $\cos(0)=1$), and symmetry across the x-axis. Observing the options:

• The first graph (selected) correctly shows the cusp at $(0,0)$ and the furthest point at $x = 14$ on the positive x-axis.
• The other graphs are either oriented along the y-axis (sine-based) or have the cusp facing the wrong direction.

Answer

The curve $r = 7(1 + \cos(\theta))$ is a cardioid symmetric about the x-axis. The total area enclosed by this curve is calculated as $\frac{147\pi}{2}$ square units.

Explanation

1. Identify the curve type and symmetry The equation $r = a(1 + \cos(\theta))$ represents a cardioid. Because it uses $\cos(\theta)$, it is symmetric with respect to the polar axis (the x-axis). Since the cosine function has a period of $2\pi$, we integrate from $\theta = 0$ to $\theta = 2\pi$ to find the total area.

2. Select the polar area formula The area $A$ in polar coordinates is given by the integral of the squared radius function. $A = \int_{\alpha}^{\beta} \frac{1}{2} r^2 \, d\theta$ This formula calculates the sum of infinitesimal circular sectors from the origin to the curve.

3. Set up the integral Substitute $r = 7(1 + \cos(\theta))$ and the interval $[0, 2\pi]$ into the formula. $A = \int_{0}^{2\pi} \frac{1}{2} [7(1 + \cos(\theta))]^2 \, d\theta$ We square the constant 7 to get 49 and move it outside the integral. $A = \frac{49}{2} \int_{0}^{2\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) \, d\theta$ Expanding the squared binomial allows us to integrate each term individually. ⚠️ This expansion is required on exams.

4. Apply trigonometric power-reduction To integrate $\cos^2(\theta)$, we must use the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$. $A = \frac{49}{2} \int_{0}^{2\pi} (1 + 2\cos(\theta) + \frac{1}{2} + \frac{1}{2}\cos(2\theta)) \, d\theta$ Combining the constants $1$ and $\frac{1}{2}$: $A = \frac{49}{2} \int_{0}^{2\pi} (\frac{3}{2} + 2\cos(\theta) + \frac{1}{2}\cos(2\theta)) \, d\theta$ This identity transforms a squared trigonometric function into a first-power function that is easier to integrate.

5. Evaluate the definite integral Now we find the antiderivative of each term. $A = \frac{49}{2} \left[ \frac{3}{2}\theta + 2\sin(\theta) + \frac{1}{4}\sin(2\theta) \right]_{0}^{2\pi}$ Evaluation at the boundaries: $A = \frac{49}{2} \left( (\frac{3}{2}(2\pi) + 2\sin(2\pi) + \frac{1}{4}\sin(4\pi)) - (0 + 0 + 0) \right)$ Since $\sin(2\pi) = 0$ and $\sin(4\pi) = 0$, the expression simplifies significantly. $A = \frac{49}{2} (3\pi)$ Multiplying the constant by the result of the integration gives the final area value.

Final Answer

The area enclosed by the curve is: $\boxed{\frac{147\pi}{2}}$

Common Mistakes

• Forgetting to square the constant: Students often forget that $r^2$ means squaring the $7$, leading to an answer that is 7 times smaller than the correct one.
• Incorrect Bounds: Integrating from $0$ to $\pi$ and forgetting to multiply by 2 (for symmetry). While you can use symmetry, you must adjust the coefficient accordingly; using the full $0$ to $2\pi$ is safer.