The discussion centers on calculating the volume of a solid of revolution using the shell method for the curve defined by y = x^3, bounded by y = 8 and x = 0. The initial attempt incorrectly applied the disc method instead of the shell method, leading to an erroneous volume calculation of 13107.2π. The correct volume, as determined by the shell method, is 768/7π. The key error was in not expressing the integrand in terms of y, which is essential for the shell method.
PREREQUISITESStudents studying calculus, particularly those focusing on volume calculations using the shell method, as well as educators looking for examples of common pitfalls in integration techniques.
What is the complete problem statement? You need to have some finite region in the plane to start with.togo said:Homework Statement
Given the following curve:
y = x^3
Use shell method and rotate around x-axis to determine the volume
togo said:Homework Equations
2pixy
The Attempt at a Solution
x(x^3) = x^4
Integrate
x^4 = 1/5 x^5
1/5(8)^5 = 6553.6
*2 = 13107.2 pi
the answer should be 768/7 pi, where did I go wrong? thanks.
For starters, if you are supposed to use shell method, revolving around the x-axis, than the integrand needs to be in terms of y, not x.togo said:the answer should be 768/7 pi, where did I go wrong? thanks.
As eumyang notes, you've used the disc method, not the shell method, but it still should produce the right answer. Your answer is wrong because you've calculated the wrong region. It is not bounded by y=0, it's bounded by y=8. Draw the bounded region in the XY plane.togo said:where did I go wrong? .