English

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

Advertisements
Advertisements

Question

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

Sum

Solution


Here, y2 = 8x and x = 2

y2 = 8(2) = 16

∴ y = ±4

Required area = `2 int_0^2 sqrt(8x)  "d"x`

= `2 xx 2sqrt(2) int_0^2 sqrt(x)  "d"x`

= `4sqrt(2) xx 2/3 [x^(3/2)]_0^2`

= `(8sqrt(2))/3 [(2)^(3/2)]`

= `(8sqrt(2))/3 xx 2sqrt(2)`

= `32/3` sq.units

Hence, the area of the region = `32/3` sq.units

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Application Of Integrals - Exercise [Page 176]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 8 Application Of Integrals
Exercise | Q 7 | Page 176

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the area of the region bounded by the curve y = sinx, the lines x=-π/2 , x=π/2 and X-axis


Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is ______.


Find the area of the region bounded by the parabola y2 = 4ax and the line x = a. 


Draw a rough sketch to indicate the region bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region.


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.


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


Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are (−1, 1), (0, 5) and (3, 2) respectively.


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


Find the area enclosed by the curve \[y = - x^2\] and the straight line x + y + 2 = 0. 


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 included between the parabolas y2 = 4x and x2 = 4y is (in square units)


The area bounded by the curve y = loge x and x-axis and the straight line x = e is ___________ .


The area of the region formed by x2 + y2 − 6x − 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is ______ .


The area bounded by the curves y = sin x between the ordinates x = 0, x = π and the x-axis is _____________ .


The area bounded by the curve y = x4 − 2x3 + x2 + 3 with x-axis and ordinates corresponding to the minima of y is _________ .


The area of the region bounded by the parabola y = x2 + 1 and the straight line x + y = 3 is given by


The area bounded by the curve y = f (x), x-axis, and the ordinates x = 1 and x = b is (b −1) sin (3b + 4). Then, f (x) is __________ .


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


Find the area of the region bound by the curves y = 6x – x2 and y = x2 – 2x 


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 y = `sqrt(16 - x^2)` and x-axis is ______.


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


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


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


Find the area of the region bounded by the ellipse `x^2/4 + y^2/9` = 1.


Find the area of the minor segment of the circle x2 + y2 = 4 cut off by the line x = 1, using integration.


Find the area of the region bounded by the curve x2 = 4y and the line x = 4y – 2.


Hence find the area bounded by the curve, y = x |x| and the coordinates x = −1 and x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×