What is the area enclosed by #r=2sin(theta+(7pi)/4) # between #theta in [pi/8,(pi)/4]#?

Answer 1

#= pi/8 -1/(2sqrt2)#

Area of a polar curve is #int_a^b 1/2 r^2 d theta#
Accordingly, in the present case it would be #int_(pi/8) ^ (pi/4) 1/2 *4sin^2 (theta + (7pi)/4) d theta#
#=int_(pi/8)^(pi/4) (1- cos(2 theta + (7pi)/2) d theta#
#[theta -1/2 sin(2theta +(7pi)/2)]_(pi/8)^(pi/4)#
#= (pi/4 -1/2 sin(pi/2 +(7pi)/2)-(pi/8 -1/2sin(pi/4+(7pi)/2)#
#(pi/4 -pi/8 +1/2 sin(4pi -pi/4))#
#=pi/8 +1/2 sin (- pi/4)#
#= pi/8 -1/(2sqrt2)#
Sign up to view the whole answer

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

Sign up with email
Answer 2

To find the area enclosed by the polar curve ( r = 2 \sin(\theta + \frac{7\pi}{4}) ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), you need to evaluate the integral of ( \frac{1}{2} r^2 ) with respect to ( \theta ) over the given interval.

The integral for the area ( A ) is given by:

[ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{1}{2} (2\sin(\theta + \frac{7\pi}{4}))^2 d\theta ]

Simplify and solve the integral:

[ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} 2\sin^2(\theta + \frac{7\pi}{4}) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} 2\sin^2(\theta + \frac{7\pi}{4}) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (1 - \cos(2(\theta + \frac{7\pi}{4}))) d\theta ] [ A = \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (1 - \cos(2\theta + \frac{7\pi}{2})) d\theta ]

Apply the integral properties:

[ A = \left[\theta - \frac{1}{2}\sin(2\theta + \frac{7\pi}{2})\right]_{\frac{\pi}{8}}^{\frac{\pi}{4}} ]

Evaluate the integral:

[ A = \left(\frac{\pi}{4} - \frac{1}{2}\sin(\frac{\pi}{2} + \frac{7\pi}{2})\right) - \left(\frac{\pi}{8} - \frac{1}{2}\sin(\pi + \frac{7\pi}{2})\right) ]

[ A = \left(\frac{\pi}{4} - \frac{1}{2}\right) - \left(\frac{\pi}{8} - \frac{1}{2}\sin(\frac{15\pi}{2})\right) ]

[ A = \frac{\pi}{4} - \frac{1}{2} - \frac{\pi}{8} ]

[ A = \frac{\pi}{8} - \frac{1}{2} ]

Therefore, the area enclosed by the polar curve between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ) is ( \frac{\pi}{8} - \frac{1}{2} ).

Sign up to view the whole answer

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

Sign up with email
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.

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.

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7