disc and shells problems

mr.burger

New member
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

Senior Member
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

Elite Member
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

New member
do you get 23.6 pi for the first and 5.6 pi for the second?

mr.burger

New member
do you get 23.6 pi for the first and 11.2 pi for the second?

soroban

Elite Member
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.)