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If is the area of the common region that lies inside both the given ellipses, then the value of is .
0.74 to 0.76
In polar form, 4x2+y2=1 is r2= 14 2 + 2 ; for 0 4 it is the inner boundary.
By symmetry about the axes and y=x, =8 0 /4 12r2 d =4 0 /4 2 d 4+ 2 .
=4 12 [ -1 2 ]0 /4=2 -1 12.
= 2 121- 14= 43 = 34=0.75.
Answer: 0.75
Let R denote the set of all real numbers. Then the area of the region \(x,y) R R:x>0, y> 1x, 5x-4y-1>0, 4x+4y-17<0 \ is
Let S denote the locus of the mid-points of those chords of the parabola y2=x, such that the area of the region enclosed between the parabola and the chord is 43. Let R denote the region lying in the first quadrant, enclosed by the parabola y2=x, the curve S, and the lines x=1 and x=4. Then which of the following statements is (are) TRUE?
Let S= \(x,y) R R: x 0, y 0, y2 4x, y2 12-2x and 3y+ 8 x 5 8 \. If the area of the region S is 2, then is equal to