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What is the area enclosed by #r=-sin(theta+(11pi)/8) -theta/4# between #theta in [0,(pi)/2]#?

Answer 1

Emma Aldridge

#A=pi/8+sqrt2/8+frac{pi^3}{768}-pi/8cos(15pi/8) + 1/4sin(15pi/8) - 1/4sin(11pi/8)#

#approx .3823#

#r=-sin(theta+frac{11pi}{8})-theta/4#

Area inside polar curves: #A=1/2int_(theta_1)^(theta_2)r^2 d theta#

#A=1/2int_(0)^(pi/2)(-sin(theta+(11pi)/8)-theta/4)^2 d theta#

(If this problem allows a calculator, use a graphing calculator here)
Expand what's inside the integral:
#1/2int_0^(pi/2)(sin^2(theta+(11pi)/8)+theta/2 sin(theta+(11pi)/8)+theta^2/16) d theta#

#=pi/8+sqrt2/8+frac{pi^3}{768}-pi/8cos(15pi/8) + 1/4sin(15pi/8) - 1/4sin(11pi/8)#

#approx .3823#

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Answer 2

Paisley Johnson

To find the area enclosed by the curve ( r = -\sin\left(\theta + \frac{11\pi}{8}\right) - \frac{\theta}{4} ) between ( \theta = 0 ) and ( \theta = \frac{\pi}{2} ), you need to integrate ( \frac{1}{2} r^2 ) with respect to ( \theta ) over the given interval. The integral will yield the desired area.

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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.