What is the question?
Can you find
Roots are really causing me some grief tonight.
Same scenario... given the following:
I keep running myself into these corners. I can plug this into my calculator and come up with:
But I'm not exactly sure how to get there. In addition, this does not look particularly useful for the integration. So, I must have made some sort of error along the way.
If you continue from
then under the square root can be factored to
Now, the graph alternates from positive to negative, so be careful with your limits,
to avoid getting a result of zero, or subtracting any part of the area.
You are correct to look for the integral of the modulus.
However, you need to locate the x-axis crossing points of your graph
in order to evaluate the area between the resulting function and the x-axis
and if it is symmetrical.
gives the area between the curve and the x-axis.
You may also express
Plugging that into my surface are formula yields:
Now, to finish this up I would like to substitute:
If I understand this correctly... I want to change my terminals of integration to ensure that I do not try to "add" a negative portion of the curve to a positive portion. How does the logic work for this?
is one complete revolution. During this revolution the function crosses the x-axis twice. Is there some rule of thumb for ensuring that I am always dealing with a positive value?
I think a value of would accomplish what I want.