How do you find the area the region of the common interior of #r=a(1+costheta), r=asintheta#?

Answer 1

#=(pi/2-1)a^2# areal units.

The points of intersection are given by

#r = a sin theta = a ( 1 + cos theta )#, giving #sin theta - cos theta = 1#
Reorganized, #sin (theta-pi/4)=1/sqrt2= sin (pi/4)#.
#theta = 0 and pi/2#, for one period #[0, 2pi].#

The area comprises one part bounded by

#theta = 0, r = a sintheta and theta = pi/2#

and the second part bounded by

#theta =pi/2, r = a (1+costheta) and theta = pi#.

Now, the area is

#1/2int r^2 d theta#, for the first area,
# + 1/2 int r^2 d theta#, for the second area.
#=a^2/2( int sin^2 theta d theta#, with #theta# from 0 to #pi/2#
#+int (1+costheta)^2 d theta#), with #theta# from #pi/2# to #pi#.
#=a^2/2(int 1/2(1-cos 2theta) d theta#, with #theta# from #0# to #pi/2#
#+ int(1+2 cos theta+1/2(1+cos 2theta ) d theta)#,
with #theta# from #pi/2 to pi#
#=a^2/2(1/2[theta-1/2 sin 2theta]#, between #theta=0 and pi/2#,
# +[theta+2sin theta+1/2(theta+1/2sin2theta)]#,
between #theta =pi/2 and pi#)
#=a^2/2([ pi/4]+[(pi-pi/2)-2+1/2(pi-pi/2)])#
#=a^2/2(pi-2)#
#=(pi/2-1)a^2#

graph{(x^2+y^2-y)(x^2+y^2-sqrt(x^2+y^2)-x)=0 [-4.034, 4.044, -2.02, 2.018]}

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Answer 2

To find the area of the region that is the common interior of ( r = a(1 + \cos \theta) ) and ( r = a \sin \theta ), you need to determine the points of intersection between the two curves by solving the equations simultaneously. Once you have the points of intersection, you can integrate to find the area enclosed by the curves.

First, set the equations equal to each other:

[ a(1 + \cos \theta) = a \sin \theta ]

Then solve for ( \theta ):

[ 1 + \cos \theta = \sin \theta ] [ 1 = \sin \theta - \cos \theta ] [ 1 = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \theta - \frac{1}{\sqrt{2}} \cos \theta \right) ] [ 1 = \sqrt{2} \sin \left( \theta - \frac{\pi}{4} \right) ]

Now, ( \sin \left( \theta - \frac{\pi}{4} \right) = 1 ) when ( \theta - \frac{\pi}{4} = \frac{\pi}{2} ).

So, ( \theta = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4} ).

Now that we have the point of intersection, we can integrate to find the area enclosed by the curves. The formula for the area enclosed by a polar curve is given by:

[ A = \frac{1}{2} \int_{\alpha}^{\beta} (f(\theta))^2 d\theta ]

Where ( f(\theta) ) is the radial function, and ( \alpha ) and ( \beta ) are the angles corresponding to the points of intersection.

Thus, the area is:

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

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

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

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (a^2 + 2a^2\cos \theta + a^2\cos^2 \theta - a^2(1 - \cos^2 \theta)) d\theta ]

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (2a^2\cos \theta + 2a^2\cos^2 \theta) d\theta ]

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2a^2 \cos \theta (1 + \cos \theta) d\theta ]

[ A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2a^2 \cos \theta d\theta + \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} 2a^2 \cos^2 \theta d\theta ]

[ A = \frac{1}{2} \left[ a^2 \sin \theta + \frac{2a^2}{2} (\theta + \frac{\sin 2\theta}{2}) \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} ]

[ A = \frac{1}{2} \left[ a^2 \sin \theta + a^2 \left( \theta + \frac{\sin 2\theta}{2} \right) \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} ]

[ A = \frac{1}{2} \left[ a^2 \left( \sin \frac{3\pi}{4} + \frac{3\pi}{4} + \frac{\sin \frac{3\pi}{2}}{2} \right) - a^2 \left( \sin \frac{\pi}{4} + \frac{\pi}{4} + \frac{\sin \frac{\pi}{2}}{2} \right) \right] ]

[ A = \frac{1}{2} \left[ a^2 \left( \frac{\sqrt{2}}{2} + \frac{3\pi}{4} - \frac{\sqrt{2}}{2} - \frac{\pi}{4} + \frac{1}{2} \right) \right] ]

[ A = \frac{1}{2} \left[ a^2 \left( \frac{3\pi}{4} - \frac{\pi}{4} + \frac{1}{2} \right) \right] ]

[ A = \frac{1}{2} \left[ a^2 \left( \frac{\pi}{2} + \frac{1}{2} \right) \right] ]

[ A = \frac{\pi a^2}{4} + \frac{a^2}{4} ]

[ \boxed{ A = \frac{\pi a^2}{4} + \frac{a^2}{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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