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) \[\pi \int\limits_{-1}^{2} (2+x)^2 - (1+(4-x^2))^2 \]

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

dx's

- amistre64

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

washer method you have to make sure your integrating and "adjusting" according to the axis of roataion as well

- 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

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

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

- anonymous

Using the washer method, that is.

- amistre64

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

- amistre64

|dw:1327958145327:dw|

- amistre64

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

- amistre64

washer is just a double disc

- TuringTest

I meant shell*

- TuringTest

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

- amistre64

thats better :)

- 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

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

- amistre64

if its a heaight, define it from the base axis

- amistre64

f-1(y) but yes

- amistre64

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

- amistre64

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

- amistre64

|dw:1327958415125:dw|

- 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

dont forget the dy if thats the case

- anonymous

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

- amistre64

y=4xâˆ’x^2
y=2âˆ’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

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

- anonymous

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

- amistre64

that picture doesnt match your equations that you posted

- 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

lol, its right everywhere except where its wrong :)

- anonymous

Hahah

- amistre64

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

- 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

Negative answer?

- amistre64

yep, so does position matter too much?

- anonymous

Nope! :) Of course, some of the grade will be setting the equation up properly, I'm sure. But I see your point!

- 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

Hahaha, exploding sun.

- 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

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

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

- amistre64

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

- 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

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

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

- 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

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

- 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

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

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

- amistre64

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

- anonymous

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

- amistre64

forgot me delimiters :)

- amistre64

one thing at time ...

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

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

- 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

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

BRB, sir!

- 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

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

- 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

ahhh, now brb! Going to be late :)

- amistre64

left and right; but yes

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.