find the volume of the solid generated by revolving the region r bounded by the graphs of the given equations about the y-axis.
x^2+y^2=1
x=1
y=1
I see a cylinder of radius 1 and height 1, containing a half sphere.
so volume of your generated solid
= π(1^1)(1) - (1/2)(4/3)π(1^3)
= π - (2/3)π
= π/3
same question by Calculus
using "washers"
outer radius = r1 = 1 --->r1^2 = 1
inner radius = r2 = √(1-y^2) --->r2^2 = 1-y^2
Volume = π∫(1 - (1-y^2) dy from y = 0 to 1
= π∫y^2 dy from y = 0 to 1
= π[ y^3/3] from 0 to 1
= π(1/3 - 0) = π/3 , same as above
To find the volume of the solid generated by revolving the region R bounded by the graphs of the given equations about the y-axis, you can use the method of cylindrical shells.
Step 1: Draw the region R bounded by the graphs of the equations x^2 + y^2 = 1, x = 1, and y = 1.
Step 2: Note that the region R is a quarter of a circle of radius 1 centered at the origin, with a rectangle of width 1 and height 1 attached to it.
Step 3: To apply the method of cylindrical shells, imagine slicing the solid into thin cylindrical shells that are infinitely small in height. The radius of each cylindrical shell will be the corresponding x-value on the circle, and the height will be the difference between the upper and lower y-values at that x.
Step 4: The volume of each cylindrical shell is given by the formula V = 2πrh, where r is the radius and h is the height.
Step 5: Integrate the volumes of all the cylindrical shells to get the total volume of the solid. Since the region R is symmetric about the y-axis, we only need to consider the right half of the region.
Step 6: Set up the integral. The radius (r) is equal to the x-coordinate of the point on the circle, which is √(1 - y^2). The height (h) is the difference between the upper and lower y-values, which is 1 - y.
Step 7: The integral to find the volume is ∫[0,1] 2π(√(1 - y^2))(1 - y) dy.
Step 8: Evaluate the integral to find the volume.