Calculus II - Volume of Solids - Setup Question \[y = 4x-x^2\] \[y=2-x\] Let D be the region bound by the curves above. Setup an integral representing the volume of the solid obtained by revolving R about each of the following lines; a) the x-axis b) the line y=5 c) the line y = -1 d) the line x = -1 e) the line x = 3 I have worked all five using the "washer method" and would like to get some feedback on the answers; a) \[\pi \int\limits_{-1}^{2} (4-x^2)^2 - (2-x)^2 \] b) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (5-(4-x^2))^2 \] c) \[\pi \int\limits_{-1}^{2} (2+x)^2 - (1+(4-x^2))^2 \]

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Calculus II - Volume of Solids - Setup Question \[y = 4x-x^2\] \[y=2-x\] Let D be the region bound by the curves above. Setup an integral representing the volume of the solid obtained by revolving R about each of the following lines; a) the x-axis b) the line y=5 c) the line y = -1 d) the line x = -1 e) the line x = 3 I have worked all five using the "washer method" and would like to get some feedback on the answers; a) \[\pi \int\limits_{-1}^{2} (4-x^2)^2 - (2-x)^2 \] b) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (5-(4-x^2))^2 \] c) \[\pi \int\limits_{-1}^{2} (2+x)^2 - (1+(4-x^2))^2 \]

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d) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (1+(4-x^2))^2 \] e) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (3-(4-x^2))^2 \]
dx's
im assuming its the area bound by the intersection of these functions

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washer method you have to make sure your integrating and "adjusting" according to the axis of roataion as well
a function viewed from the x axis doesnt look the same as a function viewed from the y axis; it becomes its inverse function
Yes, bound by the intersection. Shoot, adjusting. I also tried this for D (shell method); d) \[2\pi \int\limits_{-1}^{2} (x+1)[(4-x^2)-(2-x)]\]
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 :)
and look into your function from the way the method is looking at it; if its a radius define it from the radial axis
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? :)
y=4x−x^2 y=2−x ---------- -2+5x-x^2 = 0 x^2 -5x +2 = 0 x = 5/2 +- sqrt(17)/2 you sure your intercepts are right? Let D be the region bound by the curves above. Setup an integral representing the volume of the solid obtained by revolving R about each of the following lines; a) the x-axis
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
Ugh. I attached an X. y = 4 + x^2 Really sorry about that, but it shouldn't have effected much other then making everything wrong :)
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
you want washer rotate about the x axis for a \[\pi\int_{-1}^{2}\ (4-x^2)^2-(2-x)^2 dx\] do you know what happens when you get the top and bottom in the wrong place?
Negative answer?
yep, so does position matter too much?
Nope! :) Of course, some of the grade will be setting the equation up properly, I'm sure. But I see your point!
good, cause all the texts I see waste so much time trying to tell you to set up top on top and bottom on bottom or else the sun will explode or some nonesense. the difference between a 2 and a -2 is just drop the negative :)
Hahaha, exploding sun.
So A and B are looking good, except for dx and dy respectively? Then for C I need to change to F(y) and D E should look good?
b) the line y=5 this is the same as the xaxis; xaxis IS y=0 y=5 is just another parallel xaxis; what I do is subtract 5 from the eqs to set it back up to y=0
y = 4-x^2 -5 -5 ---------- y = -1-x^2 same eq, different position
y =2-x -5 -5 ------- y = -3-x
Damn, that's a great idea. So that prevents you from having to subtract it from 5. If I had not done that it is correct, though?
Can you do that with any Y or X = ? question. Just get the equation to F(y) or F(x) and add the x = or y = ?
you swapped top and bottom on your B and forgot to -5 from one of the eqs
yes, but with left and right shifts you have to modify the (x) perse say x^2 is measuered from x=5 instead of x=0 to move x^2 to the left by 5 we do: (x+5)^2
it just about knowing how to move the graph around to a good vantage point
Yeah, I am TERRIBLE at visualizing/setting up these graphs. Trying to get better, which is why I am on here constantly asking for setup help. Everytime I leave this website though I feel more and more confident.
so B, washered about y=5; our integration hasnt shifted left and right so x to x is still good; -1 to 2 \[\pi \int_{-1}^{2} (4-x^2-5)^2-(2-x-5)^2 dx\]
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.
d) the line x = -1 our radiuseses are now look from the y axis out so we have to invert the equations to proper perspectives
I've got to go afk for 15 minutes (moving from Library to Work) but I will be right back! If you are not here, thank you SO MUCH! You have given me a MUCH greater understanding of these concepts and I truly feel like i could tackle almost any washer problem with minimal difficulty!
BRB, sir!
y = 4 - x^2 y-4 =x^2 4-y = x^2 sqrt(4-y) = x y = 2-x y-2 = -x 2-y = x ok; and im sure they want inner outer set up "correctly" as well
but yes, to adjust the eqs for x=-1; we +1 to our eqs now
And then make sure the inner and outer equations are right. When doing inner/outer with respect to Y I assume we have to regraph the new equations and evaluate which is top and bottom?
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 :)

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