मराठी

Find the area of the region bounded by the parabolas y2 = 6x and x2 = 6y. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the region bounded by the parabolas y2 = 6x and x2 = 6y.

बेरीज

उत्तर

The intersecting points of the given parabolas are obtained by solving these equations for x and y

Which are 0(0, 0) and (6, 6).

Hence Area OABC = `int_0^6 (sqrt(6x) - x^2/6) "d"x`

= `|2sqrt(6)  x^(3/2)/3 - x^3/18|_0^6`

= `2sqrt(6)  (6)^(3/2)/3 - (6)^3/18`

= 12 sq.units

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Application Of Integrals - Solved Examples [पृष्ठ १७१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 8 Application Of Integrals
Solved Examples | Q 4 | पृष्ठ १७१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the area of the region lying in the first quandrant bounded by the curve y2= 4x, X axis and the lines x = 1, x = 4


Using integration, find the area of the region bounded by the line y − 1 = x, the x − axis and the ordinates x= −2 and x = 3.


Draw a rough sketch of the graph of the function y = 2 \[\sqrt{1 - x^2}\] , x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.


Using integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + | x + 1 |, x = −2, x = 3, y = 0.


Sketch the graph y = | x + 1 |. Evaluate\[\int\limits_{- 4}^2 \left| x + 1 \right| dx\]. What does the value of this integral represent on the graph?


Draw a rough sketch of the curve \[y = \frac{x}{\pi} + 2 \sin^2 x\] and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.


Find the area bounded by the curve y = 4 − x2 and the lines y = 0, y = 3.


Find the area of the region bounded by \[y = \sqrt{x}, x = 2y + 3\]  in the first quadrant and x-axis.


Find the area bounded by the parabola y = 2 − x2 and the straight line y + x = 0.


Find the area of the region bounded by y = | x − 1 | and y = 1.


Using integration find the area of the region bounded by the curves \[y = \sqrt{4 - x^2}, x^2 + y^2 - 4x = 0\] and the x-axis.


Find the area of the region between the parabola x = 4y − y2 and the line x = 2y − 3.


The area of the region \[\left\{ \left( x, y \right) : x^2 + y^2 \leq 1 \leq x + y \right\}\] is __________ .


The area of the region (in square units) bounded by the curve x2 = 4y, line x = 2 and x-axis is


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


Find the area of the region bounded by the parabola y2 = 2x and the straight line x – y = 4.


The area of the region bounded by the curve y = x2 and the line y = 16 ______.


Find the area of the region included between y2 = 9x and y = x


Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2


Find the area of region bounded by the line x = 2 and the parabola y2 = 8x


Using integration, find the area of the region bounded by the line 2y = 5x + 7, x- axis and the lines x = 2 and x = 8.


Draw a rough sketch of the region {(x, y) : y2 ≤ 6ax and x 2 + y2 ≤ 16a2}. Also find the area of the region sketched using method of integration.


The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is ______.


The curve x = t2 + t + 1,y = t2 – t + 1 represents


Let f(x) be a continuous function such that the area bounded by the curve y = f(x), x-axis and the lines x = 0 and x = a is `a^2/2 + a/2 sin a + pi/2 cos a`, then `f(pi/2)` =


The region bounded by the curves `x = 1/2, x = 2, y = log x` and `y = 2^x`, then the area of this region, is


Find the area of the region bounded by the curve `y = x^2 + 2, y = x, x = 0` and `x = 3`


Make a rough sketch of the region {(x, y): 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 2} and find the area of the region using integration.


The area of the region bounded by the parabola (y – 2)2 = (x – 1), the tangent to it at the point whose ordinate is 3 and the x-axis is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×