1. ## Area bounded by...

Area of region bounded by the graphs of x = y^2 and x = y+ 2

graph the area bounded by y= x^2 + 1 and y = -x + 3
find the volume of the solid formed by revolving the area about the x-axis.

harder ones i assume?
graph the area bounded by x = y^2 + 1 and x = 3

graph the area bounded by x-axis, y-axis and y = e -2x in the 1st quadrant fidn teh area?

graph the area bounded by y= 1 x=4 and y = radical x..

use disc or washer method?
use integration to find the volume.

im so clueless on how to start.. i haven't been to class for a while due to surgery on my knee and missed out a lot and we are having finals soon and im so clueless.. any help would be very much appreciated thanks guys!

"Area Between" or "Bounded by" ...

Find the intersections and decide which is farther from the pertinent axis.

x = y^2 vs. x = y+2

Intersections:

y^2 = y + 2
y^2 - y - 2 = 0
(y-2)(y+1)=0
y = 2 or y = -1

Which is farther from the Y-Axis?

y = 0 is in the Range

0^2 = 0 and 0+2 = 2, so the linear piece is farther away.

$\int_{-1}^{2}(y+2)-(y^{2})\;dy$

3. yeah i guess yur right...

anyway.. can someone help out with the other problems im trying to do it and i seem to get stuck in the middle.

4. Originally Posted by CalculusDUMMIE

graph the area bounded by y= x^2 + 1 and y = -x + 3

do you know how to graph a curve ??

first as TKHunny said you should find the point of intersection

$x^2+1=-x+3$

[tex]x^2+x-2=0[tex]

$by..quadratic..equation$

$\frac{-b\mp\sqrt{b^2-4ac}}{2a}.......for......ax^2+bx+c=0....$
you know it right
you will find the points is

$\frac{-1+\sqrt{9}}{2}....and...\frac{-1-\sqrt{9}}{2}$

$the...point...of...intersection...(1)..(-2)$

do you know how to graph y=x^2 if you know to how graph it just bush it up one unit like the graph below the second one is easier

and add y=-x+3 so you will have this