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)

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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)

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)

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

. . . V . = . (pi) ∫<sub>1 to 2</sub> (e<sup>x</sup>)<sup>2</sup> dx1) 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)

The two curves intersect at (2,8).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 "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.

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

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

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

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

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

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 ?