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Let R be the region bounded by the graphs of y=sin(x) and y=1/2 as shown in the figure above.?
a. Find the area of R.
b. Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is revolved around the x-axis.
c. The region R is the base of a solid. For each x in region R, the cross sections of this solid, perpendicular to the x-axis, are semi-circles. Write, but do not evaluate, an integral expression that gives the volume of the solid.
1 Answer
- az_lenderLv 74 weeks ago
The area is the integral from x = pi/6 to x = 5pi/6 of
[sin(x) - (1/2)] dx.
The indefinite integral is -cos(x) - (1/2)x;
plug in 5pi/6 and pi/6 to obtain
sqrt(3)/2 - 5pi/12 + sqrt(3)/2 + pi/12
= sqrt(3) - pi/3.
That's about 0.7, does it look plausible? Yes it does.
(b) I would use the "washers" method.
Area of a washer is pi[sin^2(x) - 1/4].
Integrate from x = pi/6 to 5pi/6
pi[sin^2(x) - 1/4] dx.
(c) At a particular x, the area of the desired semicircle is
(1/2)*pi*[sin(x)/2 - 1/4]^2.
Integrate from x = pi/6 to 5pi/6
(1/2)*pi*[sin(x)/2 - 1/4]^2 dx.