Find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1

We have to find area enclosed by ellipse. Since Ellipse is symmetrical about both x - axis and y - axis. ∴ Area of ellipse = 4 × Area of O A B = 4 × ∫ 0 a y d x We know that, x 2 a 2 + y 2 b 2 = 1 ⇒ y 2 b 2 = 1 - x 2 a 2 = a 2 - x 2 a 2 ⇒ y 2 = b 2 a 2 a 2 - x 2 ⇒ y = ± b 2 a 2 a 2 - x 2 ⇒ y = ± b a a 2 - x 2 Since O A B is in first quadrant, value of y is positive. ∴ y = b a a 2 - x 2 Area of ellipse = 4 × ∫ 0 a y d x = 4 ∫ 0 a b a a 2 - x 2 d x = 4 b a ∫ 0 a a 2 - x 2 d x It is of form a 2 - x 2 d x = 1 2 x a 2 - x 2 + a 2 2 sin - 1 x a + c = 4 b a x 2 a 2 - x 2 + a 2 2 sin - 1 x a 0 a = 4 b a a 2 a 2 - a 2 + a 2 2 sin - 1 a a - 0 2 a 2 - 0 + a 2 2 sin - 1 0 = 4 b a 0 + a 2 2 sin - 1 1 - 0 - 0 = 4 b a × a 2 2 × π 2 = π a b Hence, the obtained area of the ellipse is π a b sq.units

Find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1

We have to find area enclosed by ellipse. Since Ellipse is symmetrical about both x - axis and y - axis. ∴ Area of ellipse = 4 × Area of O A B = 4 × ∫ 0 a y d x We know that, x 2 a 2 + y 2 b 2 = 1 ⇒ y 2 b 2 = 1 - x 2 a 2 = a 2 - x 2 a 2 ⇒ y 2 = b 2 a 2 a 2 - x 2 ⇒ y = ± b 2 a 2 a 2 - x 2 ⇒ y = ± b a a 2 - x 2 Since O A B is in first quadrant, value of y is positive. ∴ y = b a a 2 - x 2 Area of ellipse = 4 × ∫ 0 a y d x = 4 ∫ 0 a b a a 2 - x 2 d x = 4 b a ∫ 0 a a 2 - x 2 d x It is of form a 2 - x 2 d x = 1 2 x a 2 - x 2 + a 2 2 sin - 1 x a + c = 4 b a x 2 a 2 - x 2 + a 2 2 sin - 1 x a 0 a = 4 b a a 2 a 2 - a 2 + a 2 2 sin - 1 a a - 0 2 a 2 - 0 + a 2 2 sin - 1 0 = 4 b a 0 + a 2 2 sin - 1 1 - 0 - 0 = 4 b a × a 2 2 × π 2 = π a b Hence, the obtained area of the ellipse is π a b sq.units

MEDIUM

Earn 100

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Important Questions on Application of Integrals

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in sq. unit) of the region described by

A = \{(x, y) : x^{2} + y^{2} \leq 1 and y^{2} \leq 1 - x\}

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

Consider two straight lines, each of which is tangent to both the circle

x^{2} + y^{2} = \frac{1}{2}

and the parabola

y^{2} = 4 x

. Let these lines intersect at the point

Q

. Consider the ellipse whose center is at the origin

O (0, 0)

and whose semi-major axis is

O Q

. If the length of the minor axis of this ellipse is

\sqrt{2}

, then which of the following statement(s) is (are) TRUE?

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in square units) of the region bounded by the curves

y + 2 x^{2} = 0

and

y + 3 x^{2} = 1

, is equal to

EASY

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area of region bounded by the lines

y = m x, x = 1

and

x = 2

and the

x

-axis is

7.5

sq. units, then

m

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

If the area enclosed between the curves

y = k x^{2}

and

x = k y^{2}, (k > 0),

1 s q . u n i t .

Then

k

EASY

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area enclosed between the parabolas

y^{2} = 16 x

and

x^{2} = 16 y

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in square units) bounded by the curves $y = \sqrt{x}, 2 y - x + 3 = 0$ , $X$ -axis and lying in the first quadrant is

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area of the region bounded by the lines

y = 2 x + 1, y = 3 x + 1

and

x = 4

EASY

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in sq. units) of the region bounded by the parabola,

y = x^{2} + 2

and the lines,

y = x + 1, x = 0

and

x = 3

, is

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area of the region

A = {(x, y) : 0 \leq y \leq x |x| + 1 and - 1 \leq x \leq 1}

in sq. units, is

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

If the area of the region bounded by the curves,

y = x^{2}, y = \frac{1}{x}

and the lines

y = 0

and

x = t (t > 1)

1 sq . unit

, then

t

is equal to

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

Area of the region

{(x, y) \in R^{2} : y \geq \sqrt{|x + 3|},

5 y \leq x + 9 \leq 15}

is equal to

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area of the region bounded by the curve

y = |x^{3} - 4 x^{2} + 3 x|

and the

x

-axis,

0 \leq x \leq 3,

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The parabola

y^{2} = 4 x + 1

divides the disc

x^{2} + y^{2} \leq 1

into regions with areas

A_{1}

and

A_{2} .

Then

|A_{1} - A_{2}|

equals

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in sq. units) of the region

\{(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y a n d y \leq 1 + \sqrt{x}\}

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in sq. units) of the smaller portion enclosed between the curves,

x^{2} + y^{2} = 4

and

y^{2} = 3 x

, is:

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area( in sq. units) bounded by

y = \sin^{2} x, x = \frac{π}{2}

and

x = π

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

A farmer

F_{1}

has a land in the shape of a triangle with vertices at

P (0, 0), Q (1, 1)

and

R (2, 0) .

From this land, a neighbouring farmer

F_{2}

takes away the region which lies between the side

P Q

and a curve of the form

y = x^{n}, n > 1

. If the area of the region taken away by the farmer

F_{2}

is exactly

30%

of the area of

∆ P Q R,

then the value of

n

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

Let

I = \int_{a}^{b} (x^{4} - 2 x^{2}) d x .

I

is minimum then the ordered pair

(a, b)

HARD

Mathematics>Integral Calculus>Application of Integrals>Area under Simple Curves

The area (in sq. units) of the region described by

[(x, y) : y^{2} \leq 2 x and y \geq 4 x - 1]

Find the area enclosed by the ellipse x2a2+y2b2=1

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$