Then you get:
Find the surface area of the solid of revolution over the interval
What I tried:
Plug into formula for Surface Area of a solid of revolution about the x-axis:
Convert limits: when , when
Plug in , evaluate at limits, simplify.
I don't seem to be getting the answer that's in my book.
How come you didn't have to convert the limits? do you mind evaluating this for me? I know it's really simple...I can get this far but when I evaluate I don't get the same answer as in the book:Originally Posted by ThePerfectHacker
I know that they factored...but don't quite exactly see how they got there. When I evaluate I get huge numbers. Thanks
This is how I taught my students, since they get confused with negatives and all that.
2 seperate things.
1.1)1]1}First: Evaluate it for upper and stop.
2.2)2]2}Second: Evaluate it for lower and stop.
That is it now subtract them. The reason why I prefer it like this is because it elimanates much effort with working it all out in one long line which confuses many students.
Okay, do you remember radicals?
That nasty exponent means,
Thus, the first part gives,
Okay, this is the problem are you sure that the limits are 0 and 2. They should be 1 and 2 to give the book's answer.
You see how to do it? Do you want to do the second part. I just really do not want to do it because of all the time it takes.
Now it makes sense. And let me continue.Originally Posted by c_323_h
Now you do step 2 you evaluate it for 0 only:
Okay, do you know that,
for any ?
Then, you are left with,
In my other post I did the first equaluation
In this post I did the second equaluation.
Fractions, common denominator good,
ok. i understand that you used 2 and 0 as the limits.
general question on definite intergrals: when you use u-substitution, do you always convert the limits, or can you treat the definite integral as an indefinite integral and evaluate the original limits at the end? which would be 0 and 2 in this case