Question

The x or y coordinate of the centroid
of a quadrant of a circular area of
radius r is

Answer 1

Given :  a quadrant of a circular area of radius r

To Find  : x or y coordinate of the centroid

Solution:

Assuming radius = r  and  center at origin   and 1st Quadrant

x² + y² = r²

=> y = √r² - x²

A = (1/4)πr²

(1/4)πr² x cord  =   $\int\limits^r_0 {x\sqrt{r^2-x^2} } \, dx$

$\int\limits^r_0 {x\sqrt{r^2-x^2} } \, dx = \frac{-1}{2} \int\limits^r_0 {(-2x)\sqrt{r^2-x^2} } \, dx$

r² - x²  = t   =>  -2xdx = dt

x = 0  => t = r²    and  x= r  => t = 0

$=\frac{-1}{2} \int\limits^0_{r^2} { \sqrt{t} } \, dt$

$=\frac{-1}{2} \frac{2}{3} [t^\frac{3}{2} ]_{r^2}^0$

= r³/3

(1/4)πr² x cord =  r³/3

=> x cord  =  4r/3π

Similarly y cord =  4r/3π

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Answer 2

Answer:

x or y coordinate of the centroid

Solution:

Assuming radius = r and center at origin and 1st Quadrant

x² + y² = r²

=> y = √r² - x²

A = (1/4)πr²

(1/4)πr² x cord = \int\limits^r_0 {x\sqrt{r^2-x^2} } \, dx

0

∫

r

x

r

2

−x

2

dx

\int\limits^r_0 {x\sqrt{r^2-x^2} } \, dx = \frac{-1}{2} \int\limits^r_0 {(-2x)\sqrt{r^2-x^2} } \, dx

0

∫

r

x

r

2

−x

2

dx=

2

−1

0

∫

r

(−2x)

r

2

−x

2

dx

r² - x² = t => -2xdx = dt

x = 0 => t = r² and x= r => t = 0

=\frac{-1}{2} \int\limits^0_{r^2} { \sqrt{t} } \, dt=

2

−1

r

2

∫

0

t

dt

=\frac{-1}{2} \frac{2}{3} [t^\frac{3}{2} ]_{r^2}^0=

2

−1

3

2

[t

2

3

]

r

2

0

= r³/3

(1/4)πr² x cord = r³/3

=> x cord = 4r/3π

Similarly y cord = 4r/3π