# Thread: Graph the area bounded by....

1. ## Graph the area bounded by....

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

find the volume of the solid formed by revolving teh area bout the x-axis

h=____?

DISC or washer method?

use integration to find the volume...

i was trying this problem and.. i gotten

AREA= pi r ^2

v= intergral from a to b A dx

two intergrals.. 0 to -2.. and 0 to 1

PI( R^2 Outer - r^2 in)

i had some help from a friend but im still utterly confused and i have my finals tommorow...

i really need help from you guys.. thanks alot! willing to be pressing alot of thank buttons tonight ;P

2. Originally Posted by CalculusDUMMIE
graph the area bounded by x = y^2 + 1 and x = -x +3

find the volume of the solid formed by revolving teh area bout the x-axis
h=____?
DISC or washer method?
use integration to find the volume...
x = y^2 + 1

and
x= -y + 3 (you may have a typo in your posting)

intersect at:

-y+3 = y^2 + 1
y^2 + y - 2 = 0
(y+2)(y-1) = 0

The points of intersection are (5, -2) and (2, 1).

Draw a picture to decide the method, the area consists of regions in the first quadrant and in the fourth.

While you could use a combination of disk/washer methods, I would recommend shell:

x_left = y^2 + 1
x_right = -y + 3
Height = x_right - x_left =

circumference = 2 pi distance = 2 pi y as y ranges from -2 to 1.

So Volume = Integral (2pi y Height) dy with interval of y values -2 to 1.

Good luck!!