Find the volume of the material to make a cup rotating the area between y=x+1 and y=2x^2 around x axis. x positive.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the volume of the cup created by rotating the area between y = x + 1 and y = 2x^2 around the positive x- axis.

The point of intersection of y = x + 1 and y = 2x^2 is

2x^2 = x + 1

=> 2x^2 - x - 1 = 0

=> 2x^2 - 2x + x - 1 = 0

=> 2x(x - 1) + 1(x - 1) = 0

=> (2x + 1)(x - 1) = 0

=> x = -1/2 and x = 1.

As we only have to consider the positive x values we take it from x = 0 to x = 1.

The volume that we are trying to obtain is given by the difference of the volumes enveloped by the two curves.

=>pi*Int[(x+1)^2 - (2x^2)^2 dx] , x = 0 to x = 1

=> pi*Int [ x^2 + 2x + 1 - 4x^4 dx] , x = 0 to x = 1

=> pi*(x^3/3 + 2*x^2/2 + x - 4x^5/5), x = 0 to x = 1

=> pi*(1/3 + 1 + 1 - 4/5)

=> pi*( 23/15)

The required volume is (23/15)*pi cube units.

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We only know the lower limit of integration, namely x = 0. To determine the upper limit of integration, we'll have to calculate the intercepting point of the curves.

x + 1 = 2x^2

We'll subtract x + 1 both sides:

2x^2 - x - 1 = 0

We'll apply quadratic formula:

x1 = [1 + sqrt(1 + 8)]/4

x1 = (1+3)/4

x1 = 1

x2 = -1/2

Since x has to be positive, we'll reject x2 = -1/2. The upper limit of integration is x = 1.

We'll determine the volume:

V = pi*Int[(x+1)^2 - (2x^2)^2]dx

We'll expand the square:

V = pi*Int (x^2 + 2x + 1 - 4x^4)dx

V = pi*(x^3/3 + 2*x^2/2 + x - 4x^5/5), from x = 0 to x = 1

V = pi[(1/3 + 1 + 1 - 4/5) - 0]

V = 23*pi/15 cube units.

The volume of the material requested  for making the cup is V = 23*pi/15 cube units.

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