InterviewSolution
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InterviewSolution
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The area of the region included between the curves `x^2+y^2=a^2 and sqrt|x|+sqrt|y|=sqrta(a > 0)` is |
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Answer» Given curves are ` sqrt(|x|) + sqrt(|y|) = sqrta " " … (i)` and `x^(2) + y^(2) = a^(2) " "…. (ii)` Curve (ii) is a circle . Now we have to draw the graph of curve (i) (1) Curve (i) cuts x-axis at (-a,0) and (a,0) and y-axis at (0,-a) and (0,a) (2) Curve (i) is symmetrical about x-axis as well as y-axis (3) In first quadrant from (i) we have `sqrtx + sqrty = sqrta implies (dy)/(dx) = - (sqrty)/(sqrtx) lt 0` `implies ` y is decreasing (4) `(dy)/(dx) = - ((sqrta - sqrtx)/(sqrtx)) = 1 - (sqrta)/(sqrtx)` `therefore (d^(2)y)/(dx^(2)) = (1)/(2) sqrta . x^(-(3)/(2))= (sqrta)/(2.x^((3)/(2))) gt 0` `therefore` Curve is convex downward . Now , required area = 4[shaded area in the first quadrant] =` 4[(pi a^(2))/(4) - underset(0)overset(a)(int) (sqrta - sqrtx)^(2) dx ] = 4[(pi a^(2))/(4) - underset(0)overset(a)(int)(a + x - 2sqrta sqrtx) dx`] `= 4 {(pia^(2))/(4) - [ax + (x^(2))/(2) - (4)/(3) sqrta x^((3)/(2)) ]_(0)^(a)}` `= (pi - (2)/(3)) a^(2)` sq. units |
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