disc and shells problems

mr.burger

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Jun 12, 2005
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1) The area bounded by y= e^(x) , y=0 , x=1 and x=2 is rotated around the x-axis. Find its volume. (you must use disks)


2) The area bounded by y=x^(3) , y=x^(2) +4 and x=0 is rotated about the y-axis. Find its volume. (you may use shells or disks)
 

Gene

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Oct 8, 2003
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dv = pi*e^(x)^2dx =
pi*e^(2x)dx
v = pi*Integ<sub>1 to 2</sub>e^(2x)dx =
pi/2*e^(2x)]<sub>1 to 2</sub> =
pi/2*(e^4-e^2)

By shells:
dv=2*pi*x*(x^2+4-x^3)dx =
2*pi*(x^3+4x-x^4)dx
v = 2*pi*Integ<sub>0 to 2</sub>(x^3+4x-x^4)dx =
2*pi*((x^4)/4+2(x^2)-(x^5/5))]<sub>0 to 2</sub> =
2*pi*((2^4)/4 + 2*(2^2)-(2^5)/5)
 

soroban

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Jan 28, 2005
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Hello, mr.burger!

These problems are pretty straight-forward.
Are you familiar with the Volume-of-Revolutions formulas!


Disks: . V . = . (pi) <sub>a to b</sub> (radius)<sup>2</sup> dx

Shells: . V . = . 2(pi) <sub>a to b</sub> (radius)(height) dx


1) The area bounded by y= e<sup>x</sup> , y=0 , x=1 and x=2 is rotated around the x-axis.

Find its volume. (you must use disks)
. . . V . = . (pi) <sub>1 to 2</sub> (e<sup>x</sup>)<sup>2</sup> dx


2) The area bounded by y=x<sup>3</sup> , y=x<sup>2</sup> + 4 and x = 0 is rotated about the y-axis.

Find its volume. (you may use shells or disks)
The two curves intersect at (2,8).
The "radius" is x.
The "height" is: .(x<sup>2</sup> + 4) - x<sup>3</sup>

. . . V . = . 2(pi) <sub>0 to 2</sub> x(x<sup>2</sup> + 4 - x<sup>3</sup>) dx


The set-up is 90% of the work . . . the rest is "only" integrating and evaluating.
 

mr.burger

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Jun 12, 2005
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do you get 23.6 pi for the first and 5.6 pi for the second?
 

mr.burger

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Jun 12, 2005
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do you get 23.6 pi for the first and 11.2 pi for the second?
 

soroban

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Jan 28, 2005
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Hello, mr.burger!

Oh-oh . . . I got different answers . . .

1) The area bounded by y= e<sup>x</sup> , y=0 , x=1 and x=2 is rotated around the x-axis.

Find its volume. (you must use disks)
V . = . (pi) (e<sup>x</sup>)<sup>2</sup> dx . [1 to 2]

V . = . (pi) e<sup>2x</sup> dx . = . (pi/2) e<sup>2x</sup> . [1 to 2]

V . = . (pi/2)(e<sup>4</sup> - e<sup>2</sup>) . .74.16


2) The area bounded by y=x<sup>3</sup> , y=x<sup>2</sup> + 4 and x = 0 is rotated about the y-axis.

Find its volume. (you may use shells or disks)
V . = . 2(pi) x(x<sup>2</sup> + 4 - x<sup>3</sup>) dx . [0 to 2]

V . = . 2(pi)∫ (4x + x<sup>3</sup> - x<sup>4</sup>) dx . = . 2(pi)[2x<sup>2</sup> + x<sup>4</sup>/4 - x<sup>5</sup>/5] .[0 to 2]

V . = . 2(pi) [(2·2<sup>2</sup> + 2<sup>4</sup>/4 - 2<sup>5</sup>/5) - (2·0<sup>2</sup> + 0<sup>4</sup>/4 - 0<sup>5</sup>/5)]

V . = . 2(pi)(68/5) . . 85.45

(And, as always, please check my work.)
 

Gene

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Oct 8, 2003
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Burger, yes your answers are correct.
Soroban,
We agree on the first. We agree as to the equation on the second but I get 35.19 60-32 = 28, not 68.
I have a hard time because WebTV doesn't like the codes you use. Integral, sqrt, theta... all come out as ?
 

tkhunny

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Apr 12, 2005
Messages
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First, agreed, ½*pi*e^2*(e^2 - 1) apprx 74.156
Second, (56/5)*pi apprx 35.186
 
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