Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

# Hard Calculus Problem?

Find the area if the region enclosed by the curve SQUAREROOT (x) + SQUAREROOT (y) = 1 and x=0 and y=0

Relevance
• Hemant
Lv 7

Curve : √x + √y = 1 ..... (1)

y = 0 ... => ... √x + 0 = 1 ... => ... x = 1

The req'd region is bounded by the ordinates

at x = 0 and x = 1, and the X-axis. .................... (2)

..................................................................................................................................................

Also, from (1),

√y = 1 - √x

(√y)² = ( 1 - √x )²

y = 1 - 2·√x + x ............... (3)

.................................................................................................................................................

From (2) and (3), the req'd area is

= ∫ y dx .... on [0,1]

= ∫ ( 1 - 2√x + x ) dx .... on [0,1]

= [ x - ( 4·x·√x / 3 ) + ( x² / 2 ) ] .... on [0,1]

= [ 1 - ( 4/3 ) + ( 1/2 ) ] - [ 0 ]

= ( 6 - 8 + 3 ) / 6

= ( 1/6 ) sq. units ........................... Ans.

.................................................................................................................................................

Happy To Help !

..................................................................................................................................................