Asked by Louis
The region in the first quadrant bounded by the x-axis, the line x = ฯ, and the curve y = cos(cos(x)) is rotated about the x-axis. What is the volume of the generated solid?
a. 1.921
b. 3.782
c. 6.040
d. 8.130
a. 1.921
b. 3.782
c. 6.040
d. 8.130
Answers
Answered by
Steve
think of the volume as a stack of discs of thickness dx, and you have
v = ?[0,?] ?r^2 dx
where r=y=cos(cos(x))
v = ?[0,?] ?cos^2(cosx) dx
This is not an elementary integral, so some numeric method is needed. However, using some symmetry,
v = 2?[0,?/2] ?cos^2(cosx) dx
and you can see that the volume is close to that of a truncated cone.
http://www.wolframalpha.com/input/?i=cos(cos(x))
The curve is close to the line
y = cos(1) + (1-cos(1))/(?/2) x
= 0.54 + 0.29x
So, the peak of the cone, which has been removed has a base radius of 0.54 and a height of 1.862
That makes the semi-volume nearly
?/3 (1^2)(1.862+?/2) - ?/3 (0.54^2)(1.862) = 3.02
So the whole volume is 6.04
This can be checked here:
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,%CF%80%5D+%CF%80cos%5E2(cosx)+dx
v = ?[0,?] ?r^2 dx
where r=y=cos(cos(x))
v = ?[0,?] ?cos^2(cosx) dx
This is not an elementary integral, so some numeric method is needed. However, using some symmetry,
v = 2?[0,?/2] ?cos^2(cosx) dx
and you can see that the volume is close to that of a truncated cone.
http://www.wolframalpha.com/input/?i=cos(cos(x))
The curve is close to the line
y = cos(1) + (1-cos(1))/(?/2) x
= 0.54 + 0.29x
So, the peak of the cone, which has been removed has a base radius of 0.54 and a height of 1.862
That makes the semi-volume nearly
?/3 (1^2)(1.862+?/2) - ?/3 (0.54^2)(1.862) = 3.02
So the whole volume is 6.04
This can be checked here:
http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0,%CF%80%5D+%CF%80cos%5E2(cosx)+dx