Advertisements
Advertisements
प्रश्न
Find the area of the smaller region bounded by the curves `x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1, using integration.
Advertisements
उत्तर
`x^2/25 + y^2/16` = 1 and `x/5 + y/4` = 1
`\implies` y = `4/5 sqrt(25 - x^2)`
The points of intersection of the given curve and line are A(0, 4) and B(5, 0).
Area of shaded region = Area of ellipse in I quadrant – Area of triangle ΔOAB
= `int_0^5 (4/5 sqrt(25 - x^2))dx - 1/2 xx 5 xx 4`
= `4/5 [x/2 sqrt(25 - x^2) + 25/2 sin^-1 x/5]_0^5 - 10 ...[∵ int (sqrt(a^2 - x^2))dx = x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1 x/a]`
= `4/5 [0 + 25/2 sin^-1 1] - 10`
= `10 xx π/2 - 10`
= (5π – 10) sq. units.
APPEARS IN
संबंधित प्रश्न
Using the method of integration, find the area of the triangular region whose vertices are (2, -2), (4, 3) and (1, 2).
Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is ______.
Find the equation of a curve passing through the point (0, 2), given that the sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at that point by 5
Using integration, find the area of the region bounded between the line x = 2 and the parabola y2 = 8x.
Find the area under the curve y = \[\sqrt{6x + 4}\] above x-axis from x = 0 to x = 2. Draw a sketch of curve also.
Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area of the region enclosed by it, using integration.
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.
Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x =\[\frac{\pi}{3}\] are in the ratio 2 : 3.
Find the area of the region bounded by the curve \[x = a t^2 , y = 2\text{ at }\]between the ordinates corresponding t = 1 and t = 2.
Find the area of the region between the circles x2 + y2 = 4 and (x − 2)2 + y2 = 4.
Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.
Draw a rough sketch of the region {(x, y) : y2 ≤ 5x, 5x2 + 5y2 ≤ 36} and find the area enclosed by the region using method of integration.
Prove that the area common to the two parabolas y = 2x2 and y = x2 + 4 is \[\frac{32}{3}\] sq. units.
Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 2), (1, 5) and (3, 4).
Find the area of the region bounded by the parabola y2 = 2x + 1 and the line x − y − 1 = 0.
If the area enclosed by the parabolas y2 = 16ax and x2 = 16ay, a > 0 is \[\frac{1024}{3}\] square units, find the value of a.
The area bounded by y = 2 − x2 and x + y = 0 is _________ .
The area bounded by the parabola y2 = 4ax and x2 = 4ay is ___________ .
The area bounded by the parabola y2 = 4ax, latusrectum and x-axis is ___________ .
Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3, is
The area of the region bounded by the curve y = x2 + x, x-axis and the line x = 2 and x = 5 is equal to ______.
Find the area of region bounded by the line x = 2 and the parabola y2 = 8x
Find the area of the region bounded by y = `sqrt(x)` and y = x.
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 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 P(x) be a real polynomial of degree 3 which vanishes at x = –3. Let P(x) have local minima at x = 1, local maxima at x = –1 and `int_-1^1 P(x)dx` = 18, then the sum of all the coefficients of the polynomial P(x) is equal to ______.
Hence find the area bounded by the curve, y = x |x| and the coordinates x = −1 and x = 1.
