why dont you do it horizontally?
solve for x in terms of y..
(1)
(2)
get the intersection of that 2 functions, and that would be
then,
Okay, so I'm almost finished my webwork assignment, but 3 problems still evade me. I'll say what they are and show what I have done so far.
1. FInd the area of the region enclosed by the curves 2y=3sqrt(x), y=5 and 2y+4x=7.
2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=3+(1/x^2), y=3, x=4 and x=8 about the x-axis.
So for 1, I drew a sketch based on what it looked like on my graphing calculator. Nothing too fancy, but just enough so I would know to divide my almost triangular shape into two intervals.
I found the intercepts to be (going from left to right), x=-3/4, 1 and 100/9. I found -3/4 by substituting 5 for y into 2y+4x=7. I found x=1 by setting 3sqrt(x)=7-4x, squaring both sides and solving the quadratic, but I suspect that this may be my mistake. And I found x=100/9 by substituting y=5 into y=(3sqrt(x))/2.
I did it in two parts.
Afinal=A1+a2
A1= integral(-3/4 to 1) (5-((-4x+7)/2))dx
I eventually got 0.8125 for this part.
A2=integral (1 to 100/9) (5-((3sqrt(x))/2)
I got 18.51851852 for this part.
When added, I got a final answer of 19.33101852.
I have no idea what I'm doing wrong, but I'm sure it must be something really little and annoying.