Why addition? Integral problem

[Jbreezy]

Hello, I can't seem to paste this in. So here is the link.
http://www.calcchat.com/book/Calculus-ETF-5e/
It is chapter 7, section 1, question 17.
Why are you adding ∫ [(x+2) +√4 -x] dx + ∫ 2(√4 -x) dx
?
I'm confused for 2 reasons. Why can you not just evaluate from -5 to 4? ( you will see the bounds I don't know how to put them on my integral here)
Why must you evaluate from -5 to 0 then from 0 to 4?
I understand the last term ∫ 2(√4 -x) dx this is 2 times because half is below the x axis.
But the first part...∫ [(x+2) +√4 -x] dx
Why would you add them. I can't seem to visualize why this would be. It seems I would find the area under x+2 then subtract that from the area under √4 -x.
Why not ??
Thanks,
J

 

[Mark44]

Hello, I can't seem to paste this in. So here is the link.
http://www.calcchat.com/book/Calculus-ETF-5e/
It is chapter 7, section 1, question 17.
Why are you adding ∫ [(x+2) +√4 -x] dx + ∫ 2(√4 -x) dx
?

They are really subtracting something. What is the equation of the lower half of that parabola?

I'm confused for 2 reasons. Why can you not just evaluate from -5 to 4? ( you will see the bounds I don't know how to put them on my integral here)

Because the formula for the typical area element changes at 0. Between -5 and 0, the typical area element is (y_upper - y_lower)Δx. After that, y_upper is not a value on that line.

Why must you evaluate from -5 to 0 then from 0 to 4?
I understand the last term ∫ 2(√4 -x) dx this is 2 times because half is below the x axis.
But the first part...∫ [(x+2) +√4 -x] dx
Why would you add them. I can't seem to visualize why this would be. It seems I would find the area under x+2 then subtract that from the area under √4 -x.
Why not ??
Thanks,
J

 

[Jbreezy]

They are really subtracting something. What is the equation of the lower half of that parabola?

Yeah Yeah I see. It is a negative.-

thx

 

[Jbreezy]

Similar question. http://www.calcchat.com/book/Calculus-ETF-5e/
part b they want you to rotate it about the y-axis and in part c about the line x = 3.
I don't understand this difference in writing for part b... 3^2 - (y^2)^2

And in part c they write (3-y)^2 I don't get it.
It is chapter 7 section 2 question 11.
Thanks

 

[Mark44]

Seeing that you also started a new thread for this new problem (which is the right thing to do), I am closing this thread.