Calculate the volume of the solid of revolution created by rotating the curve y

Question

Calculate the volume of the solid of revolution created by rotating the curve y = 3 + 6 exp (-2 x) about the x-axis, for x between 3 and 5. Volume : The equation of a circle of radius r, centered at the origin (0,0), is given by r2 = x2 + y2 Rearrange this equation to find a formula for y in terms of x and r. (Take the positive root.) Equation: y = What solid of revolution is swept out if this curve is rotated around the x axis, and x is allowed to vary between -rand r (You do not need to enter this answer into WebAssign.) Suppose we wanted to set up the following integral so that V gives the volume of a sphere of radius r V = f(x) dx What would a, b and f(x) be? a = (WebAssign note: remember that you enter pi as pi ) Carry out the integration, and calculate the value of V in terms of r. V =

Explanation / Answer

Plot y = 3 + 6e-2x with graphing calculator.

The solid of revolution is going to resemble a cylinder of radius 3 and height 2.

The differential volume is a circle of area y2 and thickness dx.

V = y2 dx

V = (3 + 6e-2x)2 dx

Expand and differentiate of use calculator.

V = 18.87 (a)

x2 + y2 = r2

y2 = r2 - x2

y = (r2 - x2)

y is the equation of the upper half of a circle centered at the origin. Rotating about the x-axis gives a hollow sphere of radius r.

Rotating the area between the semicircle and the x-axis between x = - r and x = 4 gives a solid sphere of radius r.

The differential volume is a circle of area y2 and thickness dx. The limits of integration are x = - r to x = r.

V = y2 dx

V = (r2 - x2) dx

a = - r

b = r

f(x) = (r2 - x2)

V = [r2x - (1/3)x3], evaluated from - r to r.

We already know the volume of a sphere is (4/3)r3, so we'll know if our calculus is good or not.

Watch the signs. Negatives to odd powers are negative.

V = [r3 - (1/3)r3] - [- r3 - (1/3)(- r3)]

V = [(2/3)r3] - [- r3 + (1/3)r3]

V = [(2/3)r3] - [- (2/3)r3]

V = [(2/3)r3 + (2/3)r3]

V = (4/3)r3    (b)

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