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

Question

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 · Answer

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 · Answer

dx's

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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 · Answer

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 · Answer

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

anonymous · Answer

Using the washer method, that is.

amistre64 · Answer

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

amistre64 · Answer

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amistre64 · Answer

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

amistre64 · Answer

washer is just a double disc

turingtest · Answer

I meant shell*

turingtest · Answer

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

amistre64 · Answer

thats better :)

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

if its a heaight, define it from the base axis

amistre64 · Answer

f-1(y)  but yes

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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