Recent content by jsun2015

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...

I correct myself, I have studied disc method. Yes. yes.

I correct myself, I have studied disc method. Yes. yes.

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...