Advertisements
Advertisements
प्रश्न
Find the area of the circle x2 + y2 = 16
Advertisements
उत्तर

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
संबंधित प्रश्न
Using integration find the area of the region {(x, y) : x2+y2⩽ 2ax, y2⩾ ax, x, y ⩾ 0}.
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^2/4 + 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.
The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
Find the area of the region bounded by the parabola y = x2 and y = |x| .
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 = x4, x = 1, x = 5 and x-axis
Sketch the graph of y = |x + 3| and evaluate `int_(-6)^0 |x + 3|dx`
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis
Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (4 , 1), B (6, 6) and C (8, 4).
Find the area of the region.
{(x,y) : 0 ≤ y ≤ x2 , 0 ≤ y ≤ x + 2 ,-1 ≤ x ≤ 3} .
Find the area of the region bounded by the following curves, the X-axis and the given lines: y = `sqrt(16 - x^2)`, x = 0, x = 4
Find the area of the region bounded by the following curves, the X-axis and the given lines: 2y = 5x + 7, x = 2, x = 8
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.
Choose the correct alternative:
Area of the region bounded by y2 = 16x, x = 1 and x = 4 and the X axis, lying in the first quadrant is ______
Choose the correct alternative:
Area of the region bounded by the parabola y2 = 25x and the lines x = 5 is ______
State whether the following statement is True or False:
The area of portion lying below the X axis is negative
State whether the following statement is True or False:
The area bounded by the curve y = f(x) lies on the both sides of the X-axis is `|int_"a"^"b" "f"(x) "d"x| + |int_"b"^"c" "f"(x) "d"x|`
The area of the circle x2 + y2 = 16 is ______
The area of the region bounded by the curve y2 = 4x, the X axis and the lines x = 1 and x = 4 is ______
The area of the region bounded by the curve y2 = x and the Y axis in the first quadrant and lines y = 3 and y = 9 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
Area enclosed between the curve y2(4 - x) = x3 and line x = 4 above X-axis is ______.
The area (in sq.units) of the part of the circle x2 + y2 = 36, which is outside the parabola y2 = 9x, is ______.
Area bounded by y = sec2x, x = `π/6`, x = `π/3` and x-axis is ______.
The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is ______.
