Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the area of the region enclosed by the polar curve
r=1+cos(theta)

Find the area of the region enclosed by the polar curve $r=1+\cos (\theta)$

View step-by-step help

View step-by-step help

Find properties of hyperbolas from equations in general form

Full solution

Q. Find the area of the region enclosed by the polar curve

r=1+\cos (\theta)

Identify curve type and symmetry: Identify the type of curve and symmetry. $\newline$ The curve given is $r = 1 + \cos(\theta)$ . This is a cardioid. Since $\cos(\theta)$ is symmetric about the vertical axis, the curve is symmetric about the $\theta = 0$ line.
Set up integral for area: Set up the integral for the area. $\newline$ The area $A$ enclosed by a polar curve from $\theta = a$ to $\theta = b$ is given by $A = \frac{1}{2} \int (r^2) d\theta$ . For a full cardioid, integrate from $0$ to $2\pi$ .
Substitute curve equation into integral: Substitute the equation of the curve into the integral. $\newline$ Substitute $r = 1 + \cos(\theta)$ into the integral. So, $A = \frac{1}{2} \int_{0}^{2\pi} (1 + \cos(\theta))^2 \, d\theta$ .
Expand and simplify integrand: Expand the integrand and simplify. $\newline$ $(1 + \cos(\theta))^2 = 1 + 2\cos(\theta) + \cos^2(\theta)$ . We know the integral of $\cos(\theta)$ over a full period is $0$ and the integral of $\cos^2(\theta)$ over a full period is $\pi$ . So, $A = \frac{1}{2} \int_{0}^{2\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) d\theta = \frac{1}{2} [2\pi + 0 + \pi]$ .
Calculate final area: Calculate the final area. $\newline$ $A = \frac{1}{2} [2\pi + \pi] = \frac{1}{2} [3\pi] = \frac{3\pi}{2}$ .

More problems from Find properties of hyperbolas from equations in general form

Question

Write the equation in standard form for the hyperbola

x^2 - y^2 = 64

\newline

Posted 2 months ago

Question

What is the center of the hyperbola

4x^2 - y^2 - 100 = 0

\newline

(\_,\_)

Posted 2 months ago

Question

What is the center of the hyperbola

(\frac{x^2}{144}) - (\frac{y^2}{81}) = 1

\newline

Posted 2 months ago

Question

What is the length of the transverse axis of the hyperbola

(\frac{x^2}{64}) - (\frac{y^2}{36}) = 1

\newline

Posted 2 months ago

Question

Write the equation in standard form for the hyperbola

y^2 - x^2 = 81

\newline

Posted 3 months ago

Question

Write the equation in standard form for the hyperbola

x^2 - 4y^2 = 100

\newline

Posted 3 months ago

Question

Write the equation in standard form for the hyperbola

9y^2 - x^2 = 36

\newline

Posted 3 months ago

Question

Write the equation in standard form for the hyperbola

4x^2 - y^2 = 16

\newline

Posted 3 months ago

Question

Write the equation in standard form for the hyperbola

y^2 - 9x^2 = 144

\newline

Posted 3 months ago

Question

y=f(x+3)

is shown. For which value of

x

, rounded to the nearest whole number, must

f(x)=0

Posted 25 days ago