Posted by **bob** on Saturday, January 31, 2009 at 4:01pm.

Compute the area of the region in the fi…rst quadrant bounded

on the left by the curve y = sqrt(x), on the right by the curve y = 6 - x,

and below by the curve y = 1.

- calculus -
**bobpursley**, Saturday, January 31, 2009 at 4:49pm
well, looking at the curves, The area will be (x2-x1) dy or (6-y - y^2 )dy ...check that.. with the limits from y=1 to ymax, or

ymax occurs when

x are the same, or

y^2=6-y

y^2+y-6=0

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

so the first quadrant ymax is 2

check that.

## Answer This Question

## Related Questions

- calculus - Find the volume of the solid generated by revolving the region about ...
- calculus - A region is bounded in the second quadrant by the curve y = ln(1–x), ...
- calculus - Find the volume of the solid generated by revolving the region about ...
- Calculus 2 - The volume of the solid generated by revolving the region about the...
- Calculus - Find the volume of the solid generated by revolving the following ...
- calculus - Which has more area, the region in the first quadrant enclosed by the...
- Calculus - I'm having trouble with the following problem: Find the volume of the...
- Calculus - 1. Find the area of the region bounded by the curves and lines y=e^x ...
- Calculus AB: Area Between Curves - Hello! I'm having trouble understanding how I...
- Calculus - The curve y = 11x - 24 - x^2 cuts the x-axis at points A and B, and ...

More Related Questions