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A curve is given by b yy={(sqrt(4-x^2)),...

A curve is given by `b yy={(sqrt(4-x^2)),0lt=x<1sqrt((3x)),1lt=xlt=3.` Find the area lying between the curve and x-axis.

Text Solution

Verified by Experts

The correct Answer is:
`(1)/(6)(2pi-sqrt(3)+36)` sq. units
The given curve is
`y={{:(sqrt((4-x^(2)))",",0lexlt1),(sqrt((3x))",",1lexle3):}`
Obviously, the curve is the are of the circle `x^(2)+y^(2)=4" (1)"`
between `0lexlt1` and the are of parabola `y^(2)=3x" (2)"`
between `1lexle3`
Required area = shaded area
=Area OABCO+Area CBDEC
`=|int_(0)^(1)sqrt((4-x^(2)))dx|+|int_(1)^(3)sqrt(3x)dx|`
`|[(1)/(2)xsqrt((4-x^(2)))+(4)/(2)sin^(-1)((x)/(2))]_(0)^(1)|+|sqrt(3)[(2)/(3)x^(3//2)]_(1)^(3)|`
`=((sqrt(3))/(2)+(pi)/(3))+(2)/(3)(9-sqrt(3))`
`=(1)/(6)(2pi-sqrt(3)+36)` sq. units.
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