Advertisements
Advertisements
Question
Find the area of the circle x2 + y2 = 16
Advertisements
Solution
By the symmetry of the circle, required area of the circle is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 4.
Given equation of the circle is x2 + y2 = 16
∴ y2 = 16 – x2
∴ y = `+- sqrt(16 - x^2)`
∴ y = `sqrt(16 - x^2)` ......[∵ In first quadrant, y > 0]
∴ Required area = 4(area of the region OPQO)
= `4 xx int_0^4 y*"d"x`
= `4 xx int_0^4 sqrt(16 - x^2) "d"x`
= `4int_0^4 sqrt((4)^2 - x^2) "d"x`
= `4[x/2 sqrt((4)^2 - x^2) + (4)^2/2 sin^-1 (x/4)]_0^4`
= `4{[4/2 sqrt((4)^2 - (4)^2) + 16/2 sin^-1 (4/4)] - [0/2 sqrt((4)^2 - (0)^2) + 16/2 sin^-1 (0/4)]}`
= `4{[0 + 8 sin^-1 (1)] - [0 + 0]}`
= `4(8 xx pi/2)`
= 16π sq.units
APPEARS IN
RELATED QUESTIONS
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Find the area of the region bounded by the ellipse `x^2/16 + y^2/9 = 1.`
Find the area of the region in the first quadrant enclosed by x-axis, line x = `sqrt3` y and the circle x2 + y2 = 4.
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line `x = a/sqrt2`
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is ______.
Find the area under the given curve and given line:
y = x2, x = 1, x = 2 and x-axis
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).
Find the area bounded by the circle x2 + y2 = 16 and the line `sqrt3 y = x` in the first quadrant, using integration.
State whether the following is True or False :
The area bounded by the two cures y = f(x), y = g (x) and X-axis is `|int_"a"^"b" f(x)*dx - int_"b"^"a" "g"(x)*dx|`.
State whether the following is True or False :
The area bounded by the curve y = f(x), X-axis and lines x = a and x = b is `|int_"a"^"b" f(x)*dx|`.
State whether the following is True or False :
The area of the portion lying above the X-axis is positive.
The area of the region lying in the first quadrant and bounded by the curve y = 4x2, and the lines y = 2 and y = 4 is ______
The area of the region x2 = 4y, y = 1 and y = 2 and the Y axis lying in the first quadrant is ______
Find the area of the region bounded by the curve y = `sqrt(9 - x^2)`, X-axis and lines x = 0 and x = 3
Find the area of the region bounded by the curve 4y = 7x + 9, the X-axis and the lines x = 2 and x = 8
Find area of the region bounded by 2x + 4y = 10, y = 2 and y = 4 and the Y-axis lying in the first quadrant
Find the area of the region bounded by the curve x = `sqrt(25 - y^2)`, the Y-axis lying in the first quadrant and the lines y = 0 and y = 5
The area bounded by y = `27/x^3`, X-axis and the ordinates x = 1, x = 3 is ______
The area of the region bounded by the X-axis and the curves defined by y = cot x, `(pi/6 ≤ x ≤ pi/4)` is ______.
The area enclosed by the parabolas x = y2 - 1 and x = 1 - y2 is ______.
The area bounded by the X-axis, the curve y = f(x) and the lines x = 1, x = b is equal to `sqrt("b"^2 + 1) - sqrt(2)` for all b > 1, then f(x) is ______.
Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x, is:
If a2 + b2 + c2 = – 2 and f(x) = `|(1 + a^2x, (1 + b^2)x, (1 + c^2)x),((1 + a^2)x, 1 + b^2x, (1 + c^2)x),((1 + a^2)x, (1 + b^2)x, 1 + c^2x)|` then f(x) is a polynomial of degree
The slope of a tangent to the curve y = 3x2 – x + 1 at (1, 3) is ______.
Area bounded by y = sec2x, x = `π/6`, x = `π/3` and x-axis is ______.
The area bounded by the curve | x | + y = 1 and X-axis is ______.
Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0,y = 2 and y = 4.
