The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work.

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The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work.

Calculus1
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The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. http://goo.gl/jXIZD 1. Find the area of R. 2. The region R is the base of a solid. For each y, where 0<=y<=2, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 2y. Write, but do not evaluate, an integral expression that gives the volume of the solid. 3. There is a point P on the graph of f at which the line tangent to the graph of f is perpendicular to the graph of g. Find the coordinates of point P.
I calculated 22/3 for 1. Is that right? What do I do for 2 and 3?

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wow .. could you summarize ... I am hard time reading through all.
one by one.
hmm for area of region you can do \[ \int_0^4 f(x) dx + \int_4^6 g(x) dx\]
Right. I believe I've already gotten the answer for the area. Which is exactly what you gave. Which ends up being 16/3 + 2 = 22/3. I'm not sure what to do for 2 and 3.
could you draw figure of 2 ??
|dw:1362066203257:dw| Is that what it's supposed to look like? It says the height is 2y and the base is R (which I don't understand).
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.
So are you saying that it ends up being the integral from [0,4] of 2f(x)^2 + the integral from [4,6] of 2 g(x)^2 dx?
yes ... how do you find volume .. .simplify it , you will ge tthat expression \[ \iiint dV\]
Now what should I do about part 3?
Part 3 is looking for a point on f where it's perpendicular to g(x). I assume that the point must be somewhere between x=0-1. How do I find it?
the part 3 is not difficult .. just find the slope using f'(x) and equate it with m1*m2 = -1 ... my being slope of g(x)
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.
In order for that equation to work, I would need to find a way for 1/2√x = 1 so then 1 * -1 =-1 right? Well, what if 1/2√x doesn't reach one at any given moment?
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
haha ... I'll give more complete answer for 2 \[ \int_0^4 \int_0^{f(x)} \int_0^{2y} dz dy dz + \int_0^4 \int_0^{g(x)} \int_0^{2y} dz dy dz\] First case, put y=f(x) ... second case, put y=g(x)
Woops!! \[ \int_0^4 \int_0^{f(x)} \int_0^{2y} dz dy dz + \int_4^6 \int_0^{g(x)} \int_0^{2y} dz dy dz \] and simplifty it.

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