The problem involves finding the volume of a solid of revolution using the shell method, specifically for the curve defined by y = x^3, when rotated around the x-axis. The bounded region is described by the lines y = 8 and x = 0.
Some participants have pointed out potential misunderstandings regarding the method of integration and the definition of the bounded region. There is an ongoing exploration of the correct approach to take, with no explicit consensus reached yet.
Participants note that the original poster's calculations may have misinterpreted the region of integration, and there is a suggestion to visualize the bounded area in the XY plane for clarity.
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? .