Home → Calculus → Applications that Integrals → Volume that a hard of Revolution: Disks and Washers

If a an ar in the plane is revolved around a heat in the same plane, the resulting object is well-known as a heavy of revolution.

You are watching: Find the volume of the solid generated by revolving the region bounded by

For example, a solid appropriate circular cylinder deserve to be generated by revolving a rectangle. Similarly, a heavy spherical ball deserve to be produced by revolving a semi-disk.

The line around which we rotate the form is referred to as the axis that revolution.

The disc Method

The disk technique is offered when we turn a single curve (y = fleft( x ight)) approximately the (x-) (or (y-)) axis.

Suppose that (y = fleft( x ight)) is a consistent non-negative role on the interval (left< a,b ight>.)

*
Figure 1.

The volume of the solid developed by revolving the an ar bounded by the curve (y = fleft( x ight)) and also the (x-)axis between (x= a) and also (x = b) about the (x-)axis is provided by


^2dx .>

The cross ar perpendicular come the axis of change has the form of a decaying of radius (R = fleft( x ight).)

Similarly, us can find the volume the the solid once the region is bounded by the curve (x = fleft( y ight)) and also the (y-)axis in between (y = c) and also (y = d,) and is rotated around the (y-)axis.

*
Figure 2.

The resulting formula is


^2dy .>

The Washer Method

We can prolong the disk an approach to uncover the volume that a hole solid of revolution.

Assuming the the functions (fleft( x ight)) and (gleft( x ight)) are consistent and non-negative top top the interval (left< a,b ight>) and (gleft( x ight) le fleft( x ight),) take into consideration a an ar that is bounded by two curves (y = fleft( x ight)) and also (y = gleft( x ight),) in between (x = a) and (x = b.)

*
Figure 3.

The volume that the solid developed by revolving the region about the (x-)axis is


^2 - left< gleft( x ight) ight>^2 ight)dx .>

At a allude (x) ~ above the (x-)axis, a perpendicular cross section of the solid is washer-shape with the within radius (r = gleft( x ight)) and the external radius (R = fleft( x ight).)

The volume that the solid generated by revolving around the (y-)axis a an ar between the curve (x = fleft( y ight)) and (x = gleft( y ight),) where (gleft( y ight) le fleft( y ight)) and (c le y le d) is offered by the formula


^2 - left< gleft( y ight) ight>^2 ight)dy .>
*
Figure 4.

Volume the a heavy of revolution for a Parametric Curve

If a bounding curve is defined in parametric kind by the equations (x = xleft( t ight),) (y = yleft( t ight),) where the parameter (t) different from (alpha) come (eta,) climate the volume of the solid generated by revolving the curve around the (x-)axis is provided by



Respectively, as soon as the curve is rotated about the (y-)axis, the volume of the solid of revolution is equal



Volume the a solid of revolution for a Polar Curve

There are plenty of curves the are offered by a polar equation (r = rleft( heta ight).) To convert from polar coordinates (left( r, heta ight)) come Cartesian works with (left( x,y ight),) we use the well-known formulas



So we concerned the parametric kind of the curve taken into consideration in the vault section.

It is important to save in mind the the radius vector (r) also depends on the parameter ( heta.) Therefore, the derivatives (fracdxdt) and (fracdydt) room written as




Solved Problems

Click or madness a difficulty to check out the solution.


Example 1

Using the decaying method, calculate the volume the the best circular cone of height (H) and also base radius (R.)


Example 2

Find the volume that the solid obtained by rotating the sine duty between (x = 0) and also (x = pi) about the (x-)axis.


Example 3

Calculate the volume the the solid obtained by rotating the an ar bounded by the parabola (y = x^2) and the square root role (y = sqrt x) roughly the (x-)axis.


Example 4

Find the volume the the solid derived by rotating the region bounded by two parabolas around the (x-)axis.


Example 1.

Using the disk method, calculate the volume of the best circular cone of elevation (H) and also base radius (R.)


Solution.

The slant elevation of the cone is characterized by the equation:



*
Figure 5.

Hence, the volume of the cone is offered by


^2dy = pi intlimits_0^H left< R - fracRHy ight>^2dy = pi R^2intlimits_0^H left( 1 - frac2yH + fracy^2H^2 ight)dy = pi R^2left. left( H - fracy^2H + fracy^33H^2 ight) ight|_0^H = pi R^2left( cancelH -cancelH + fracH3 ight) = fracpi R^2H3.>

Example 2.

Find the volume that the solid derived by rotating the sine function between (x = 0) and also (x = pi) about the (x-)axis.


Solution.

*
Figure 6.

By the disc method,


^2dx = fracpi 2intlimits_0^pi left( 1 - cos 2x ight)dx = fracpi 2left. left( x - fracsin 2x2 ight) ight|_0^pi = fracpi 2left< left( pi - 0 ight) - left( 0 - 0 ight) ight> = fracpi ^22.>

Example 3.

Calculate the volume that the solid obtained by rotating the region bounded through the parabola (y = x^2) and also the square root duty (y = sqrt x) roughly the (x-)axis.


Solution.

*
Figure 7.

Both curves intersect at the clues (x = 0) and also (x = 1.) utilizing the washer method, us have


^2 - left< x^2 ight>^2 ight)dx = pi intlimits_0^1 left( x - x^4 ight)dx = pi left. left( fracx^22 - fracx^55 ight) ight|_0^1 = pi left( frac12 - frac15 ight) = frac3pi 10.>

Example 4.

Find the volume that the solid derived by rotating the an ar bounded by two parabolas about the (x-)axis.


Solution.

First we determine the borders (a) and (b:)



Hence the limits of integration are (a = -1,) (b = 1.) We map out the bounding an ar and the heavy of revolution:

*
Figure 8.

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Using the washer method, we discover the volume the the solid:


^2 - left< gleft( x ight) ight>^2 ight)dx = pi intlimits_ - 1^1 left( left( 3 - x^2 ight)^2 - left( x^2 + 1 ight)^2 ight)dx = pi intlimits_ - 1^1 left( left< 3 - x^2 ight>^2 - left< x^2 + 1 ight>^2 ight)dx = pi intlimits_ - 1^1 left( 8 - 8x^2 ight)dx = 8pi intlimits_ - 1^1 left( 1 - x^2 ight)dx = 8pi left. left( x - fracx^33 ight) ight|_ - 1^1 = 8pi left< left( 1 - frac13 ight) - left( - 1 + frac13 ight) ight> = 8pi cdot frac43 = frac32pi 3.>

See more problems on web page 2.


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