Advertisements
Advertisements
प्रश्न
Find the area between the two curves (parabolas)
y2 = 7x and x2 = 7y.
Advertisements
उत्तर
To find the point of intersection of the curves:
Equation of curve is x2 = 7y
∴ y = `x^2/7`
y2 = `x^4/49` ......(1)
Equation to second curve,
y2 = 7x ......(2)
Equating equations (1) and (2) we get
`x^4/49` = 7x
⇒ x4 = 343x
⇒ x4 – 343x = 0
⇒ x(x3 – 343) = 0
⇒ x = 0
or x3 = 343
⇒ x = 7
When x = 0, y = 0
When x = 7, y = 7
∴ The points of intersection of parabolas are (0, 0) and (7, 7).
∴ Required area, A = `|int_0^7 y_1 . dx - int_0^7 y_2. dx|`
= `|int_0^7 sqrt(7x) . dx - int_0^7 x^2/7. dx|`
= `|sqrt(7) [x^(3/2)/(3/2)]_0^7 - 1/7 [x^3/3]_0^7|`
= `|2/3 xx sqrt(7) xx 7^(3/2) - 1/21 7^3|`
= `|2/3 xx 7^2 - 1/21 xx 7^3|`
= `|2/3 xx 7^2 - 1/3 xx 7^2|`
= `7^2(2/3 - 1/3)`
= `49/3` sq.units
Hence area between the two curves is `49/3` sq.units.
APPEARS IN
संबंधित प्रश्न
Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.
Using integration find the area of the region {(x, y) : x2+y2⩽ 2ax, y2⩾ ax, x, y ⩾ 0}.
Using integration find the area of the triangle formed by positive x-axis and tangent and normal of the circle
`x^2+y^2=4 at (1, sqrt3)`
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 in the first quadrant enclosed by x-axis, line x = `sqrt3` y and the circle x2 + y2 = 4.
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
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`
Find the equation of an ellipse whose latus rectum is 8 and eccentricity is `1/3`
Draw a rough sketch and find the area bounded by the curve x2 = y and x + y = 2.
Find the area of the region bounded by the following curves, the X-axis, and the given lines:
y = `sqrt(6x + 4), x = 0, x = 2`
Find the area of the region bounded by the following curves, the X-axis and the given lines: 2y + x = 8, x = 2, x = 4
The area of the region bounded by y2 = 4x, the X-axis and the lines x = 1 and x = 4 is _______.
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:
Using the definite integration area of the circle x2 + y2 = 16 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 circle x2 + y2 = 16 is ______
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 area of the region bounded by the curve y = – 4x, the X-axis and the lines x = – 1 and x = 2
The area enclosed between the curve y = loge(x + e) and the coordinate axes 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 area included between the parabolas y2 = 4a(x +a) and y2 = 4b(x – a), b > a > 0, 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 the curves y = `"e"^(x^2)`, the x-axis and the lines x = 1, x = 2 is given to be α square units. If the area bounded by the curve y = `sqrt(ℓ "n"x)`, the x-axis and the lines x = e and x = e4 is expressed as (pe4 – qe – α), (where p and q are positive integers), then (p + q) is ______.
Area bounded by y = sec2x, x = `π/6`, x = `π/3` and x-axis is ______.
