coputer engineering

Question

Find the volume of the solid generated by revolving the region bounded by y = x 2 , y = 0 and x = 2 about the y-axis.

Answer 1

y= x²

when x=2 , y=4

x = y^1/2

^{d V = A d y}

^y=4

V = ∫ π ( r_o² - r_i² ) dy , where r_o is the outer radius of the washer =2 , r_o² =4

^y=0 and ri is the inner radius of the washer= y^1/2 , r_i² =y

^{y=4 y= 4}

= π ∫ ( 4 - y ) dy = π ( 4y - y²/2) Ι = π ( 16- 16/2 ) = 8 π

^y=0 y=0

Also you can use shell method where V = circumference * height* thickness

dV = 2 π x * x² *dx = 2 π x ³ dx

^{x=2 x=2}

V = ∫ 2 π x ³ dx = 2 π x ⁴/ 4 Ι = 2 π 2 ⁴/ 4 = 8π

^{x=0 x=0}

Answer 2

I'm assuming you mean y = 0 and y = 2 (not x = 2 because that doesn't make sense).

y = x² is a parabola with vertex (0,0). Revolving that around the y-axis produces a parabolic cup of sorts.

To find the volume we create tiny discs and add up those volumes. Each individual disc will have a base area of πr² and a height of Δx to give us a volume of πr²Δx. In our case the radius is x, so we have Volume of a disc = πx²Δx.

We integrate that from 0 to 2 so we get

∫πx²dx = πx³/3 from 0 to 2 or π(2)³/3 - π(0)³/3 = 8π/3

So the volume of the solid created by revolving y=x² about the y-axis is 8π/3

Answer 3

The volume of a shell is 2pi* Height*Radius*thickness. Use tht to define the volume in this problem as a series of shells whose height is f(x) and radius is x and the thickness is dx. Thus the differential volume dV=2pi*x²dx. Since the maximum radius is given at 2 the volume can be found by finding the integral (0-2) of 2pi*x²dx which yields 8pi