Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is ______.

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KollegeApply · Accepted Answer

Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is 4$\pi$ sq.units. Explanation: Given equation of circle is x2 + y2 = 32 $\Rightarrow$ x2 + y2 = (4sqrt(2))^2 and the line is y = x and the x-axis. Solving the two equations We have x2 + x2 = 32 $\Rightarrow$ 2x2 = 32 $\Rightarrow$ x2 = 16 $\therefore$ x = $\pm$ 4 Required area = `int_0^4 x "d"x + int_4^(4sqrt(2)) sqrt((4sqrt(2))^2 - x^2) "d"x = $\frac{1}{2}$ [x^2]_0^4 + [x/2 sqrt((4sqrt(2))^2 - x^2) + $\frac{32}{2}$ sin^-1 x/(4sqrt(2))]_4^(4/sqrt(2)) = $\frac{1}{2}$ [16 - 0] + [0 + 16 sin^-1 ((4sqrt(2))/(4sqrt(2))) - 2sqrt(32 - 16) - 16sin^-1 4/(4sqrt(2))] = 8 + [16 sin^-1 (1) - 8 - 16sin^-1 1/sqrt(2)] = 8 + 16 * pi/2 - 8 - 16 * pi/4 = 8pi - 4pi` = 4$\pi$ sq.units [Solution diagram: 251725_solution_1.png]

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