What is the area enclosed by #r=2sin(4theta+(11pi)/12) # between #theta in [pi/8,(pi)/4]#?

Answer 1

# (pi + 1)/8#

It's the red segment.

Plug and play

#A = 1/2 \int_{\theta = pi/8}^{pi/4} \ r^2(theta) \ d theta #
# = \int_{\theta = pi/8}^{pi/4} \ 2sin^2 (4theta+(11pi)/12) \ d theta#

using # cos 2A = 1 - 2 sin^2 A#

# = \int_{\theta = pi/8}^{pi/4} \ 1 - cos (8theta+(11pi)/6) \ d theta#

# = [ \ theta - 1/8 sin (8theta+(11pi)/6) ]_{\theta = pi/8}^{pi/4}#

# = [ \ pi/4 - 1/8 sin (2 pi+(11pi)/6) ] - [ \ pi/8 - 1/8 sin ( pi+(11pi)/6) ]#

# = pi/8 + 1/8 [ sin ( pi+(11pi)/6)- sin (2 pi+(11pi)/6) ]#

# = pi/8 + 1/8 [ sin ( pi) cos ((11pi)/6) + cos ( pi) sin ((11pi)/6) - sin (2 pi) cos ((11pi)/6) - cos (2 pi) sin ((11pi)/6) ]#

# = pi/8 + 1/8 [ (-1) (- 1/2) - (1)(-1/2) ]#

# = (pi + 1)/8#

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Answer 2

To find the area enclosed by the polar curve ( r = 2\sin(4\theta + \frac{11\pi}{12}) ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), you can use the formula for finding the area enclosed by a polar curve:

[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta ]

where ( r ) is the polar function and ( \theta_1 ) and ( \theta_2 ) are the limits of integration.

First, find the expression for ( r^2 ):

[ r^2 = (2\sin(4\theta + \frac{11\pi}{12}))^2 ]

Now, integrate ( r^2 ) with respect to ( \theta ) over the given interval:

[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (2\sin(4\theta + \frac{11\pi}{12}))^2 d\theta ]

After integrating, you will have the area enclosed by the given polar curve between ( \frac{\pi}{8} ) and ( \frac{\pi}{4} ).

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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.

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