1. I thought outer radius would be that because at x=1, radius =8, and at x=0 radius is 8-4ln(3-x)
2. the functions intersect at x=0 and x=1, I thought because of that there would not be a difference in changing functions for radius.
Homework Statement
1. R is the shaded region in the 1st quadrant bounded by the graph of y=4ln(3-x), the horizontal line y=6, and the vertical line x=2
Find the volume of the solid when revolved about the horizontal line y=8
2. Let R be the region in the 1st quadrant enclosed by the graphs of...
Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0
I have only studied the washer method.
Yes. Yes. I came up with my original answer because at the outer region 1= radius so we have one and at the inner region 0 = radius as (1^2-1)^2=0
I have only studied the washer method.
r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)
limits to integration
the radius of the sphere is on the region x= 0 to 1
r^2
1 represents outer radius (the larger radius)
x^2-1 represents the inner radius (smaller radius)
limits to integration
the radius of the sphere is on the region x= 0 to 1
Homework Statement
Find the volume of the solid generated by revolving the region bounded by the parabola y=x^2 and the line y=1 about the line y=1
Homework Equations
V= integral of pi*r^2 from a to b with respect to variable "x"
The Attempt at a Solution
pi(integral of 1-(x^2-1)^2...
Homework Statement
Using Washer Method: Revolve region R bounded by y=x^2 and y=x^.5 about y=-3
Homework Equations
V= integral of A(x) from a to b with respect to a variable "x"
A(x)=pi*radius^2
The Attempt at a Solution
pi(integral of (x^.5-3)^2 -(x^2)^2-3) from 0 to 1 with...