Consider the dude r = 7 cos(theta). (Round you answers to two decimal places.) F
ID: 2875821 • Letter: C
Question
Consider the dude r = 7 cos(theta). (Round you answers to two decimal places.) Find the area of the darkle. Complete the table giving the areas A of the sectors of the circle between theta = 0 and the values of theta in the table. Use a graphing utility to approximate, to two decimal places, the angles theta to which the sector of the composes 1/4, 1/2, and 3/4 of the total area of the drift. Do the Texas of part depend on the radon of the Dade? yes they depend on the radius no they do not depend on the radius.Explanation / Answer
The given circle r=7cos(theta)
a) We first find a and b
7cos(theta)=0
cos(theta)=0 then theta =pi/2,3pi/2
a=pi/2 and b=3pi/2
A=(1/2)*integration of (from a=pi/2 to 3pi/2)r^2 d(theta)
A=(1/2)*integration of (from a=pi/2 to 3pi/2)(7cos(theta))^2 d(theta)
A=(49/2)*integration of (from a=pi/2 to 3pi/2)cos(theta))^2 d(theta)
A=(49/2)*integration of (from a=pi/2 to 3pi/2)cos(theta))^2 d(theta)
A=(49/2)*integration of (from a=pi/2 to 3pi/2)[(1+cos2(theta))]/2 d(theta)
A=(49/2){(1/2)theta+(sin2(theta))/2}(from a=pi/2 to 3pi/2)
A=(49/2){[3pi/4+(sin(3pi))/2]-[pi/4+(sin(pi))/2]}
A=(49/2){(3pi/4+0)-(pi/4+0)}=(49/2)*(2pi/4)=(49*pi)/4=38.48
b) We know the area of the circle is
When theta=0 and 0.2
A=(1/2)*integration of (from theta=0 and 0.2)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 0.2) since from above (a)
A=(49/2)*[(0.2)/2+(sin(0.4))/2]=7.22
When theta =0 and 0.4
A=(1/2)*integration of (from theta=0 and 0.4)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 0.4) since from above (a)
A=(49/2)[(0.4)/2 +(sin(0.8))/2]=13.68
When theta =0 and 0.6
A=(1/2)*integration of (from theta=0 and 0.6)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 0.6) since from above (a)
A=(49/2)[(0.6)/2 +(sin(1.2))/2]=18.77
When theta =0 and 0.8
A=(1/2)*integration of (from theta=0 and 0.8)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 0.8) since from above (a)
A=(49/2)[(0.8)/2 +(sin(1.6))/2]=22.04
When theta =0 and 1.0
A=(1/2)*integration of (from theta=0 and 1.0)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 1.0) since from above (a)
A=(49/2)[(1.0)/2 +(sin(2))/2]=23.38
When theta =0 and 1.2
A=(1/2)*integration of (from theta=0 and 1.2)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 1.2) since from above (a)
A=(49/2)[(1.2)/2 +(sin(2.4))/2]=22.97
When theta =0 and 1.4
A=(1/2)*integration of (from theta=0 and 1.4)[(7cos(theta))^2 d(theta)]
A=(49/2){(1/2)theta+(sin2(theta))/2}(from theta=0 and 1.4) since from above (a)
A=(49/2)[(1.4)/2 +(sin(2.8))/2]=21.25
c) Total area of the circle is 129.31 since adding the above areas of circles from 0 to 1.4
1/4 th of the area is (1/4)*129.31=32.33
1/2 th of the area is (1/2)*129.31=64.65
3/4 th of the area is (3/4)*129.31=96.98