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
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Find the area of the region bounded by the parabola y = x2 and y = |x| .
Find the area under the given curve and given line:
y = x2, x = 1, x = 2 and x-axis
Find the area under the given curve and given line:
y = x4, x = 1, x = 5 and x-axis
Find the area of the smaller region bounded by the ellipse `x^2/a^2 + y^2/b^2 = 1` and the line `x/a + y/b = 1`
Draw a rough sketch and find the area bounded by the curve x2 = y and x + y = 2.
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 = x4, x = 1, x = 5
Find the area of the region bounded by the following curve, the X-axis and the given line:
y = 2 – x2, x = –1, x = 1
Area of the region bounded by y = x4, x = 1, x = 5 and the X-axis is _______.
State whether the following is True or False :
The area bounded by the curve x = g (y), Y-axis and bounded between the lines y = c and y = d is given by `int_"c"^"d"x*dy = int_(y = "c")^(y = "d") "g"(y)*dy`
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|`.
Choose the correct alternative:
Area of the region bounded by the curve y = x3, x = 1, x = 4 and the X-axis is ______
Choose the correct alternative:
Area of the region bounded by the curve x2 = 8y, the positive Y-axis lying in the first quadrant and the lines y = 4 and y = 9 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 ______
State whether the following statement is True or False:
The equation of the area of the circle is `x^2/"a"^2 + y^2/"b"^2` = 1
The area of the shaded region bounded by two curves y = f(x), and y = g(x) and X-axis is `int_"a"^"b" "f"(x) "d"x + int_"a"^"b" "g"(x) "d"x`
The area of the region x2 = 4y, y = 1 and y = 2 and the Y axis lying in the first quadrant is ______
`int "e"^x ((sqrt(1 - x^2) * sin^-1 x + 1)/sqrt(1 - x^2))`dx = ________.
Area bounded by the curve xy = 4, X-axis between x = 1, x = 5 is ______.
The area of the region bounded by the curve y = x IxI, X-axis and the ordinates x = 2, x = –2 is ______.
Which equation below represents a parabola that opens upward with a vertex at (0, – 5)?
The area of the region bounded by the curve y = x2, x = 0, x = 3, 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 ______.
The area bounded by the curve, y = –x, X-axis, x = 1 and x = 4 is ______.
Which strips are considered for the area bounded by \[y=f(x)\], the \[x\]-axis, and the ordinates \[x=a\] and \[x=b\]?
A curve crosses the \[x\]-axis, forming area \[A_1\] below the axis and area \[A_2\] above the axis. What is its total area?
Why is the area enclosed by \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] written as \[4\int_0^a y\,dx\]?
Which integral represents the enclosed area of the ellipse after using the positive first-quadrant value of \[y\]?
