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 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 ______.
Sketch the graph of y = |x + 3| and evaluate `int_(-6)^0 |x + 3|dx`
Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x - 2y + 12 = 0.
Find the equation of an ellipse whose latus rectum is 8 and eccentricity is `1/3`
Find the area of the region.
{(x,y) : 0 ≤ y ≤ x2 , 0 ≤ y ≤ x + 2 ,-1 ≤ x ≤ 3} .
Using integration find the area of the triangle formed by negative x-axis and tangent and normal to the circle `"x"^2 + "y"^2 = 9 "at" (-1,2sqrt2)`.
Choose the correct alternative :
Area of the region bounded by the curve x2 = y, the X-axis and the lines x = 1 and x = 3 is _______.
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.
If the curve, under consideration, is below the X-axis, then the area bounded by curve, X-axis and lines x = a, x = b is positive.
Solve the following:
Find the area of the region bounded by the curve x2 = 25y, y = 1, y = 4 and the Y-axis.
Choose the correct alternative:
Area of the region bounded by the curve y = x3, x = 1, x = 4 and the X-axis is ______
Area of the region bounded by the curve x = y2, the positive Y axis and the lines y = 1 and y = 3 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 ______
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 bounded by y2 = 25x, x = 1 and x = 2 the X axis is ______
Find the area of the region bounded by the parabola y2 = 25x and the line x = 5
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 x = `sqrt(25 - y^2)`, the Y-axis lying in the first quadrant and the lines y = 0 and y = 5
Find the area of the region bounded by the curve y = `sqrt(36 - x^2)`, the X-axis lying in the first quadrant and the lines x = 0 and x = 6
`int_0^log5 (e^xsqrt(e^x - 1))/(e^x + 3)` dx = ______
`int "e"^x ((sqrt(1 - x^2) * sin^-1 x + 1)/sqrt(1 - x^2))`dx = ________.
The area of the region bounded by the X-axis and the curves defined by y = cot x, `(pi/6 ≤ x ≤ pi/4)` is ______.
Equation of a common tangent to the circle, x2 + y2 – 6x = 0 and the parabola, y2 = 4x, is:
Area of the region bounded by y= x4, x = 1, x = 5 and the X-axis is ______.
The area (in sq. units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is ______.
If the area enclosed by y = f(x), X-axis, x = a, x = b and y = g(x), X-axis, x = a, x = b are equal, then f(x) = g(x).
