anonymous
  • anonymous
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) \[\pi \int\limits_{-1}^{2} (2+x)^2 - (1+(4-x^2))^2 \]
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
d) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (1+(4-x^2))^2 \] e) \[\pi \int\limits_{-1}^{2} (2-x)^2 - (3-(4-x^2))^2 \]
amistre64
  • amistre64
dx's
amistre64
  • amistre64
im assuming its the area bound by the intersection of these functions

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amistre64
  • amistre64
washer method you have to make sure your integrating and "adjusting" according to the axis of roataion as well
amistre64
  • amistre64
a function viewed from the x axis doesnt look the same as a function viewed from the y axis; it becomes its inverse function
anonymous
  • anonymous
Yes, bound by the intersection. Shoot, adjusting. I also tried this for D (shell method); d) \[2\pi \int\limits_{-1}^{2} (x+1)[(4-x^2)-(2-x)]\]
anonymous
  • anonymous
So anytime I am rotating about the Y-axis I need to adjust my equations to be in terms of Y?
anonymous
  • anonymous
Using the washer method, that is.
amistre64
  • amistre64
adjust your equations according to axis and method of ...sheel, disc, washer etc
amistre64
  • amistre64
|dw:1327958145327:dw|
amistre64
  • amistre64
R = f(x) r = g(x) pi (R^2 - r^2) = washer area integrated with the x axis
amistre64
  • amistre64
washer is just a double disc
TuringTest
  • TuringTest
I meant shell*
TuringTest
  • TuringTest
shell about x integrate w/respect to x disk about x integrate w/respect to y and vice-versa
amistre64
  • amistre64
thats better :)
amistre64
  • amistre64
and look into your function from the way the method is looking at it; if its a radius define it from the radial axis
anonymous
  • anonymous
Ok, so if you rotate around y axis R = f(y) r = g(y)
amistre64
  • amistre64
if its a heaight, define it from the base axis
amistre64
  • amistre64
f-1(y) but yes
amistre64
  • amistre64
inverse when you are looking at it from the other axis that the equation is defined for
amistre64
  • amistre64
y=x^2 looking at it from the yaxis is x=sqrt(y)
amistre64
  • amistre64
|dw:1327958415125:dw|
anonymous
  • anonymous
So I am assuming A is correct, but for B; \[\pi \int\limits_{-1}^{2}(y-2)^2 - (5-(\sqrt(4-y)))^2\] ?
amistre64
  • amistre64
dont forget the dy if thats the case
anonymous
  • anonymous
But tacking a dy onto the end its correct? :)
amistre64
  • amistre64
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
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=4x%E2%88%92x%5E2%3D2-x
anonymous
  • anonymous
We're actually given a picture to go along with the assignment; |dw:1327958708347:dw|
amistre64
  • amistre64
that picture doesnt match your equations that you posted
anonymous
  • anonymous
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 :)
amistre64
  • amistre64
lol, its right everywhere except where its wrong :)
anonymous
  • anonymous
Hahah
amistre64
  • amistre64
better, -1,3 and 2,0 are going to be our "limits" or at least we can keep them in mind
amistre64
  • amistre64
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?
anonymous
  • anonymous
Negative answer?
amistre64
  • amistre64
yep, so does position matter too much?
anonymous
  • anonymous
Nope! :) Of course, some of the grade will be setting the equation up properly, I'm sure. But I see your point!
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
Hahaha, exploding sun.
anonymous
  • anonymous
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?
amistre64
  • amistre64
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
amistre64
  • amistre64
y = 4-x^2 -5 -5 ---------- y = -1-x^2 same eq, different position
amistre64
  • amistre64
y =2-x -5 -5 ------- y = -3-x
anonymous
  • anonymous
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?
anonymous
  • anonymous
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 = ?
amistre64
  • amistre64
you swapped top and bottom on your B and forgot to -5 from one of the eqs
amistre64
  • amistre64
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
amistre64
  • amistre64
it just about knowing how to move the graph around to a good vantage point
anonymous
  • anonymous
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.
amistre64
  • amistre64
so B, washered about y=5; our integration hasnt shifted left and right so x to x is still good; -1 to 2 \[\pi \int_{-1}^{2} (4-x^2-5)^2-(2-x-5)^2 dx\]
amistre64
  • amistre64
c) the line y = -1 this is the same thing; but +1 to move things to the right spot
amistre64
  • amistre64
\pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx
anonymous
  • anonymous
Now when it is X = -1 we will need to get it in terms of Y and then add one?
amistre64
  • amistre64
forgot me delimiters :)
amistre64
  • amistre64
one thing at time ...
anonymous
  • anonymous
Haha, k. Trying to apply that principle to the next questions.
amistre64
  • amistre64
\[B=\pi \int_{-1}^{2} (4-x^2+1)^2-(2-x+1)^2 dx\]
amistre64
  • amistre64
that was C lol, but its a fancy looking mistake so its ok
anonymous
  • anonymous
Beautiful, got the same thing written on my notepad in front of me.
amistre64
  • amistre64
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
anonymous
  • anonymous
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!
anonymous
  • anonymous
BRB, sir!
amistre64
  • amistre64
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
amistre64
  • amistre64
but yes, to adjust the eqs for x=-1; we +1 to our eqs now
anonymous
  • anonymous
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?
anonymous
  • anonymous
ahhh, now brb! Going to be late :)
amistre64
  • amistre64
left and right; but yes
amistre64
  • amistre64
the graphs dont change; we just see them from the y axis instead of the x axis
amistre64
  • amistre64
of cours ethe math doesnt care if you use and x or y; just as long as your consistent
amistre64
  • amistre64
for D about x=-1 \[\pi\int_{y=0}^{y=3}(\sqrt{4-y}+1)^2-(2-y+1)^2dy\]
amistre64
  • amistre64
my ride is ready to leave, so thanks you for the time and good luck :)

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