Let
Then you get:
Let
Find the surface area of the solid of revolution over the interval
1.
Answer:
What I tried:
Plug into formula for Surface Area of a solid of revolution about the x-axis:
let
Convert limits: when , when
Integrate:
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.
---
Step 1)
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.
Now subtract,
Fractions, common denominator good,
Factor pi,
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