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I calculated 22/3 for 1. Is that right? What do I do for 2 and 3?

wow .. could you summarize ... I am hard time reading through all.

one by one.

hmm
for area of region you can do
\[ \int_0^4 f(x) dx + \int_4^6 g(x) dx\]

could you draw figure of 2 ??

No ... draw 3d coordinates. with x-y at flat.

? 3d?

|dw:1362066629575:dw|

What does that do for us?

|dw:1362067183705:dw|
Q 2 says that your volume is like this

Isn't it supposed to be a rectangle?

No ... just like wedge.

How is the "wedge" represented with the functions that I have?

|dw:1362067339866:dw|

So is the integral √x dx on the interval of [0,2]? I'm not quite sure what I should be seeing.

... let's ignore Region 2 for a whle.

|dw:1362067567201:dw|

You make nice drawings. ;) But I'm still not sure how to evaluate them.

for volume ... use this integral|dw:1362067702501:dw|
do same for other region.

yes ... how do you find volume .. .simplify it , you will ge tthat expression
\[ \iiint dV\]

Now what should I do about part 3?

woops!! what's up with my typing!! f'(x)*m2 = -1 and m2 is slope of g(x)

m2 = -1?

f'(x) = 1/2√x so 1/2√x * -1 = -1?

yea ... use that value to find .. x and use x to find y.

never mind. lol. X = 1/4th I think.

So now that I have x how would I get y? Plug it into the equation of f?

yes ,,,

Hurray!! THanks!! I wish I had more medals to give you. ;)

it's okay!! you are welcome :)

Found y as 1/2

yep ... that's correct!!

Yay my life is now complete!! xD