Asked by Will
find the approximate volume in cubic units of the solid created when the region under the curve y = sec(x) on the interval [0, pi / 3 ] is rotated around the x-axis.
Answers
Answered by
Reiny
Exact answer:
volume = π ∫ (sec^2 x) dx from 0 to π/3
= π [ tanx ] from 0 to π/3 , since the derivative of tanx = sec^2 x
= π( tan π/3 - tan 0)
= √3 π
In your question, we have a case where the solution to find an approximate answer is more
difficult than the actual solution above.
You could think of the volume of your solid as a series of very thin fulcrums, that requires knowing how to find the volume of a fulcrum:
http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html
perhaps splitting [0,π/3] into 3 parts might be a sufficient approximation.
I don't know at what level of math this question comes from, so I don't know what method has been suggested to you.
volume = π ∫ (sec^2 x) dx from 0 to π/3
= π [ tanx ] from 0 to π/3 , since the derivative of tanx = sec^2 x
= π( tan π/3 - tan 0)
= √3 π
In your question, we have a case where the solution to find an approximate answer is more
difficult than the actual solution above.
You could think of the volume of your solid as a series of very thin fulcrums, that requires knowing how to find the volume of a fulcrum:
http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html
perhaps splitting [0,π/3] into 3 parts might be a sufficient approximation.
I don't know at what level of math this question comes from, so I don't know what method has been suggested to you.
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