Home > Calculus > Calculating Polar Areas

What is the area under the polar curve #f(theta) = theta^2-thetasin(2theta-pi/4 ) +cos(3theta-(5pi)/4)# over #[pi/8,pi/2]#?

Answer 1

Aurora Alvarado

#A ~~ 0.65#

Area with polar coordinates

#A = int_(alpha)^(beta)(1/2)r^2d theta#

Substitute the #f(theta)# for r:

#A = (1/2)int_(pi/8)^(pi/2)(theta^2 - thetasin(2theta- pi/4) + cos(3theta - (5pi)/4))^2d theta#

Integration of the definate integral by WolframAlpha

#A ~~ 0.65#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To find the area under the polar curve ( f(\theta) = \theta^2 - \theta \sin(2\theta - \frac{\pi}{4}) + \cos(3\theta - \frac{5\pi}{4}) ) over ( [\frac{\pi}{8}, \frac{\pi}{2}] ), integrate ( \frac{1}{2} f(\theta)^2 ) with respect to ( \theta ) from ( \frac{\pi}{8} ) to ( \frac{\pi}{2} ). This gives the area enclosed by the curve.

So, the integral to compute the area is:

[ \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{2}} ( \theta^2 - \theta \sin(2\theta - \frac{\pi}{4}) + \cos(3\theta - \frac{5\pi}{4}))^2 d\theta ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.