हिंदी

Find the area of the region bounded by the curve y2 = 2x and x2 + y2 = 4x. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the region bounded by the curve y2 = 2x and x2 + y2 = 4x.

योग
Advertisements

उत्तर

Equations of the curves are given by x2 + y2 = 4x  ......(i)

And y2 = 2x   ......(ii)

⇒ x2 – 4x + y2 = 0

⇒ x2 – 4x + 4 – 4 + y2 = 0

⇒ (x – 2)2 + y2 = 4

Clearly it is the equation of a circle having its centre (2, 0) and radius 2.


Solving x2 + y2 = 4x and y2 = 2x

x2 + 2x = 4x

⇒ x2 + 2x – 4x = 0

⇒ x2 – 2x = 0

⇒ x(x – 2) = 0

∴ x = 0, 2

Area of the required region

= `2[int_0^2 sqrt(4 - (x - 2)^2)  "d"x - int_0^2 sqrt(2x)  "d"x]`

∴ Parabola and circle both are symmetrical about x-axis.

= `2[(x - 2)/2 sqrt(4 - (x - 2)^2) + 4/2 sin^-1  (x - 2)/2]_0^2 - 2*sqrt(2)* 2/3 [x^(3/2)]_0^2`

= `2[(0 + 0) - (0 + 2 sin^-1 (-1)] - (4sqrt(2))/3 [2^(3/2) - 0]`

= `-2 xx 2 * (- pi/2) - (4sqrt(2))/3 * 2sqrt(2)`

= `2pi - 16/3`

= `2(pi - 8/3)` sq.units

Hence, the required area = `2(pi - 8/3)` sq.units

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Application Of Integrals - Exercise [पृष्ठ १७७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 8 Application Of Integrals
Exercise | Q 16 | पृष्ठ १७७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the area bounded by the curve y2 = 4axx-axis and the lines x = 0 and x = a.


Using the method of integration, find the area of the triangular region whose vertices are (2, -2), (4, 3) and (1, 2).


Sketch the region bounded by the curves `y=sqrt(5-x^2)` and y=|x-1| and find its area using integration.


Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3+ 5 = 0


The area bounded by the curve y = x | x|, x-axis and the ordinates x = –1 and x = 1 is given by ______.

[Hint: y = x2 if x > 0 and y = –x2 if x < 0]


Find the area lying above the x-axis and under the parabola y = 4x − x2.


Sketch the graph of y = \[\sqrt{x + 1}\]  in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.


Draw a rough sketch of the graph of the curve \[\frac{x^2}{4} + \frac{y^2}{9} = 1\]  and evaluate the area of the region under the curve and above the x-axis.


Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.


Find the area enclosed by the curve x = 3cost, y = 2sin t.


Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.


Calculate the area of the region bounded by the parabolas y2 = x and x2 = y.


Find the area of the region common to the parabolas 4y2 = 9x and 3x2 = 16y.


Draw a rough sketch and find the area of the region bounded by the two parabolas y2 = 4x and x2 = 4y by using methods of integration.


Find the area, lying above x-axis and included between the circle x2 + y2 = 8x and the parabola y2 = 4x.


Sketch the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 1. Also, find the area of this region.


Using 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 enclosed by the parabola x2 = y and the line y = x + 2.


Find the area of the region bounded by the curve y = \[\sqrt{1 - x^2}\], line y = x and the positive x-axis.


Find the area bounded by the lines y = 4x + 5, y = 5 − x and 4y = x + 5.


If the area bounded by the parabola \[y^2 = 4ax\] and the line y = mx is \[\frac{a^2}{12}\] sq. units, then using integration, find the value of m. 

 


The area bounded by the parabola y2 = 4ax, latusrectum and x-axis is ___________ .


The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π/3 is ________ .


Area lying in first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2, is


Using integration, find the area of the region bounded by the line x – y + 2 = 0, the curve x = \[\sqrt{y}\] and y-axis.


Find the equation of the parabola with latus-rectum joining points (4, 6) and (4, -2).


Using integration, find the area of the smaller region bounded by the ellipse `"x"^2/9+"y"^2/4=1`and the line `"x"/3+"y"/2=1.`


The area of the region bounded by the curve x = y2, y-axis and the line y = 3 and y = 4 is ______.


Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.


Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2π


The area of the region bounded by the y-axis, y = cosx and y = sinx, 0 ≤ x ≤ `pi/2` is ______.


The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is ______.


Area lying in the first quadrant and bounded by the circle `x^2 + y^2 = 4` and the lines `x + 0` and `x = 2`.


The area bounded by the curve `y = x^3`, the `x`-axis and ordinates `x` = – 2 and `x` = 1


Let the curve y = y(x) be the solution of the differential equation, `("dy")/("d"x) = 2(x + 1)`. If the numerical value of area bounded by the curve y = y(x) and x-axis is `(4sqrt(8))/3`, then the value of y(1) is equal to ______.


Let f : [–2, 3] `rightarrow` [0, ∞) be a continuous function such that f(1 – x) = f(x) for all x ∈ [–2, 3]. If R1 is the numerical value of the area of the region bounded by y = f(x), x = –2, x = 3 and the axis of x and R2 = `int_-2^3 xf(x)dx`, then ______.


Sketch the region enclosed bounded by the curve, y = x |x| and the ordinates x = −1 and x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×