# Area Bounded by Curves & Volume of Solid of Revolution

Find the area of each polar region enclosed by f(theta) for a <=theta<=b

36) f(theta) = theta/pi, 0<=theta<=2pi

PLEASE SHOW EVERY STEP IN SOLVING THESE-NO COMPUTER PROGRAMS PLEASE.

4) Identify each curve as cardiode, rose(state # of petals), leminscate, limacon, circle, line or none of above.

a) r=2sin2theta

b) r^2=2cos2theta

c) r=5cos60degrees

d) r=5sin8theta

e) rtheta=3

f) r^2=9cos(2theta-pi/4)

g) r=sin3(theta+pi/6)

h) costheta=1-r

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Chapter 7

6) Integral t^2 dt/9+t^6

Use substitution to integrate certain powers of sine and cosine.

42) Integral sin^4x cox x dx

43) Integral sin^2x cox^2x dx

Find:

46)iIntegral e^x dx/1+e^x/2

50) integraldx/[(x+0.5)(SQRT 4x^2+4x)]

52) Find the area of the region bounded by the graphs of y=2x/SQRTx^2+9 and y=0 from x=0 to x=4.

54) Find the volume of the solid generated when the curve y=x(1-x^2)^1/4 from x=0 to x=1 is revolved about the x axis.

The picture given is a conical cone around the x axis going from 0 to 9 on the x axis. The highest point of the cone I y=3 and low y=-3. starts at 0 and blows up to cone size at x=9.

56) Find the volume of the solid generated when the curve x= 4th root of 4-y^2 between y=1 and y=2 is revolved about the y axis.

58) Find the arc length of the curve y= ln(cos x) on the interval [0,pi/4].

60) Find the area of the surface generated when the curve y= x^2 on the interval [0,1] is revolved about the y axis.

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(see attached)

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#### Solution Summary

Twenty problems related to Area Bounded by Curves, Volume of Solid of Revolution, Arc Length, Identifying - Cardiode, Rose (state # of petals), Leminscate, Limacon, Circle, Line are solved. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.[Editor's Note: This is one of the most comprehensive solutions I have seen, of this type.]