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dx's

im assuming its the area bound by the intersection of these functions

So anytime I am rotating about the Y-axis I need to adjust my equations to be in terms of Y?

Using the washer method, that is.

adjust your equations according to axis and method of ...sheel, disc, washer etc

|dw:1327958145327:dw|

R = f(x)
r = g(x)
pi (R^2 - r^2) = washer area integrated with the x axis

washer is just a double disc

I meant shell*

shell about x
integrate w/respect to x
disk about x
integrate w/respect to y
and vice-versa

thats better :)

Ok, so if you rotate around y axis
R = f(y)
r = g(y)

if its a heaight, define it from the base axis

f-1(y) but yes

inverse when you are looking at it from the other axis that the equation is defined for

y=x^2 looking at it from the yaxis is x=sqrt(y)

|dw:1327958415125:dw|

So I am assuming A is correct, but for B;
\[\pi \int\limits_{-1}^{2}(y-2)^2 - (5-(\sqrt(4-y)))^2\]
?

dont forget the dy if thats the case

But tacking a dy onto the end its correct? :)

http://www.wolframalpha.com/input/?i=4x%E2%88%92x%5E2%3D2-x

We're actually given a picture to go along with the assignment;
|dw:1327958708347:dw|

that picture doesnt match your equations that you posted

lol, its right everywhere except where its wrong :)

Hahah

better, -1,3 and 2,0 are going to be our "limits" or at least we can keep them in mind

Negative answer?

yep, so does position matter too much?

Hahaha, exploding sun.

y = 4-x^2
-5 -5
----------
y = -1-x^2
same eq, different position

y =2-x
-5 -5
-------
y = -3-x

you swapped top and bottom on your B and forgot to -5 from one of the eqs

it just about knowing how to move the graph around to a good vantage point

c) the line y = -1
this is the same thing; but +1 to move things to the right spot

\pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx

Now when it is X = -1 we will need to get it in terms of Y and then add one?

forgot me delimiters :)

one thing at time ...

Haha, k. Trying to apply that principle to the next questions.

\[B=\pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx\]

that was C lol, but its a fancy looking mistake so its ok

Beautiful, got the same thing written on my notepad in front of me.

BRB, sir!

but yes, to adjust the eqs for x=-1; we +1 to our eqs now

ahhh, now brb! Going to be late :)

left and right; but yes

the graphs dont change; we just see them from the y axis instead of the x axis

of cours ethe math doesnt care if you use and x or y; just as long as your consistent

for D about x=-1
\[\pi\int_{y=0}^{y=3}(\sqrt{4-y}+1)^2-(2-y+1)^2dy\]

my ride is ready to leave, so thanks you for the time and good luck :)