Oh I see what ive been doing now. When I evaluated theta at 0 I was getting -2. I forgot to multiply this by 1/2 which would have given me 3(pi)(-2+2)/2 or 3(pi)/2 for my final answer. Thank you so much for all of your help I was stuck on that problem for awhile. :smile:
Use a double integral to find the area of the region bounded by the curve r= 1+sin(theta)?
The Attempt at a Solution I can't figure out what theta is intregrated from. I've tried from -(pi)/2 -> +(pi)/2 and that doesnt work. I've also tried...
The point I picked on the line l for the parametric equation was not point (1,1,2)... So whatever point you pick on the line for your parametric equation is that your t=0 point?
If I were to pick point (1,1,2) on the line l for the parametric equation then it becomes x=1+t u=1+2t z=2+3t...
I know that the magnitude of the vector with the point c is 5. So using cos(36.7)=x/5
I get x=4 this is the length along line l. However this is in three dimensions so I thought that there was something else I had to do.
Find point d on the line l closest to the point c (1,1,7). Point c is on the end of a vector who's origin (1,1,2) is on line l. There is an imaginary line that connects point c to point d. This imaginary line is perpendicular to the line l. This problem is relating to...
I know it is supposed to be lnx however I find something peculiar. When I integrate it in wolfram alpha they give the integral as log(x). What the heck is going on here!?!
The Attempt at a Solution
Find the area between f(x)=(x-1)^3 and f(x)=(x-1) on the interval from 0 to 2.
The Attempt at a Solution Working it out Im using the top function minus the bottom function from 0 to 1 and then from 1 to 2. The graphs cross at x=0,x=1,andx=2...