Saturday

April 18, 2015

April 18, 2015

Posted by **Sophie** on Thursday, January 21, 2010 at 3:14am.

- calculus -
**Reiny**, Thursday, January 21, 2010 at 9:24amNice Question!!!

The first one we can do without Calculus, it is simply the segment between the chord from (0,1) to (1,0), that is

pi/4 - 1/2 = (pi - 2)/4

the second equation is

√x + √y = 1

or

√y = 1-√x

y = (1-√x)^2 = 1 - 2√x + x for 0 <= x <= 1

the integral of 1 - 2√x + x is x - (4/3)x^(3/2) + (1/2)x^2

and the area enclosed by √x + √y = 1 , the x-axis, and the y-axis is (from 0 to 1)

(1 - 4/3 + 1/2) - 0

= 1/6

So the area between the curve √x + √y = 1 and the line x+y=1 is

1/2 - 1/6 = 1/3

then (pi-2)/4 = .2854

1/3 = .33333

So who is bigger?

**Answer this Question**

**Related Questions**

Calculus AB...I really need help - The region in the first quadrant enclosed by ...

calculus - Sketch the region in the first quadrant enclosed by the given curves...

calculus - 1. Let R be the region in the first quadrant enclosed by the graphs ...

calculus - Find the area of the region in the first quadrant enclosed by the ...

calculus - Let R be the region in the first quadrant that is enclosed by the ...

Calc AB - What is the area of the region in the first quadrant enclosed by the ...

calculus - Find the volume of the solid generated by revolving the region about ...

AP Calculus - Let R be the first quadrant region enclosed by the graph of y= 2e...

calculus - A region is bounded in the second quadrant by the curve y = ln(1–x), ...

math - The equations y=-2X+12, y=6-X, and X=0 enclose a region in quadrant I. ...