Question

Math

Posted 3 months ago

```
The polar curve $r(\theta)=\theta \cdot \sin (2 \theta)$ is graphed for $0 \leq \theta \leq \pi$.
Let $R$ be the region in the fourth quadrant enclosed by the curve and the $x$ -axis.
Which integral represents the area of $R$ ?
Choose 1 answer:
(A) $\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$
(B) $\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$
(C) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{4} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$
(ㄷ) $\int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot \theta^{2} \cdot \sin ^{2}(2 \theta) d \theta$
```

Sign in to unlock the answer

Answer from Sia

Posted 3 months ago

Solution by Steps

step 2

The area $A$ of a region enclosed by a polar curve $r(\theta)$ from $\alpha$ to $\beta$ is given by the integral $A = \frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 d\theta$

step 3

For the region $R$ in the fourth quadrant, $\theta$ ranges from $\frac{\pi}{2}$ to $\pi$. Thus, we need to evaluate the integral $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} (\theta \cdot \sin(2\theta))^2 d\theta$

step 4

Simplifying the integrand, we get $A = \frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \theta^2 \cdot \sin^2(2\theta) d\theta$

step 5

Comparing this with the given choices, the correct integral is $D: \int_{\frac{\pi}{2}}^{\pi} \frac{1}{2} \cdot \theta^2 \cdot \sin^2(2\theta) d\theta$

D

Key Concept

Polar Area Formula

Explanation

The area of a region enclosed by a polar curve $r(\theta)$ from $\alpha$ to $\beta$ is given by $\frac{1}{2} \int_{\alpha}^{\beta} r(\theta)^2 d\theta$.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages

Strong algorithms that better know you

Early access to new release features

Study Other Question