Find the centroid of the region bounded by the given curves. y = x³, x + y = 2, y = 0

Solution:

The region bounded by y = x³, x + y = 2, and y = 0 is shown below:

Let

f(x) = 2 - x or x = 2 - y

g(x) = x³ or x = y¹/³

They intersect at (1,1)

To find the area bounded by the region we could integrate w.r.t y as shown below

A = \( \int_{0}^{1}((2-y) - y^{1/3})dy\)

= \( \left [ 2y - \dfrac{1}{2}y^{2} - \dfrac{3}{4}y^{4/3} \right]_{0}^{1} \)

= 2 - 1/2 - 3/4 - (0 - 0 - 0)

= 3/4

To find the y coordinate of the centroid

\(\bar Y\)= 1/(3/4) \( \int_{0}^{1}y((2-y)- y^{1/3})dy \)

= 4/3\( \int_{0}^{1}(2y - y^{2} - y^{4/3)})dy \)

= 4/3\( [y^{2} - \dfrac{1}{3}y^{3}-\dfrac{3}{7}y^{7/3}]_{0}^{1} \)

= 4/3[1 - 1/3 - 3/7 - (0 - 0 -0)]

= 4/3[(21 - 7 - 9)/21]

= 20/63

The x coordinate of the centroid is obtained as

\(\bar X\)= (4/3)(1/2)\( \int_{0}^{1}((2-y)^{2} - (y^{1/3})^{2}))dy \)

= (2/3)\( [4y - 2y^{2} + \dfrac{1}{3}y^{3} - \dfrac{3}{5}y^{5/3}]_{0}^{1}  \)

= (2/3)[4 - 2 + 1/3 - 3/5 - (0 - 0 + 0 - 0)]

= (2/3)[2 + 1/3 - 3/5]

= (2/3)[(30 + 5 - 9)/15]

= (2/3)[26/15]

= 52/45

Hence the coordinates of the centroid are (\(\bar X\), \(\bar Y\)) = (52/45, 20/63)

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Find the centroid of the region bounded by the given curves. y = x³, x + y = 2, y = 0

Summary:

The coordinates of the centroid are (\(\bar X\), \(\bar Y\))= (52/45, 20/63).