find the volume using shell methody=3x^2 , y=3√x

Question

use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves and lines about the x-axis. y=3x^2 ,y=3√x

Answer 1

When 3x²=3√x at the points of intersection.

x²=√x⇒x=0,1 and respectively y=0,3, that is, (0,0) and (1,3) are the intersection points.

In the shell method the x-axis is the axis of rotation so y forms the radius of the cylindrical shells and x forms the height. Each shell is a cylinder lying on its side. The height varies, starting at zero height at (0,0), then growing and shrinking back to zero at (1,3). To find the height we need x in terms of y:

x²=y/3 so x=√(y/3) and √x=y/3, so x=y²/9. The height is √(y/3)-y²/9, which is positive between y=0 and y=3; for example, at y=¾, the height is ½-¹/₁₆=⁷/₁₆.

The volume of each cylindrical shell is 2πy(√(y/3)-y²/9)dy, and the sum of all the infinitesimal shells is:

2π₀∫³y(√(y/3)-y²​​​​​​​/9)dy=

2π₀∫³(y^3/2/√3-y³/9)dy=

2π[⅖y^5/2/√3-y⁴/36]₀³=2π(18/5-9/4)=2.7π cubic units.