Question

> Find the area of the boundary ellipse??
$\frac{x {}^{2} }{16} + \: \frac{y {}^{2} }{9}$
> ᴅᴏɴ'ᴛ sᴘᴀᴍ❌❌

> ɪᴛ's ᴜʀɢᴇɴᴛ

> ᴄᴏʀʀᴇᴄᴛ ᴀɴsᴡᴇʀ = ʙʀᴀɪɴʟɪᴇsᴛ✨✨​

Answer 1

Answer:

Given:

The ellipse is,

$x {}^{2} \div 16 + y {}^{2} \div 9 = 1$

Here,

a²=16

a=4

And,b²=9

b=3

Method-1:

Area of the region bounded by the ellipse=πab

Area=π×4×3

Area=12πunits

The area of the region bounded by the ellipse=12π units.

Method-2:

The ellipse is,

x²/16+y²/9=1

y²/9=1-x²/16

y²=9/16(16-x²)

y=±3/4√16-x²

Area of the region=12πunits(Refer to the attachment)

Step-by-step explanation:

Hope it helps you........

Answer 2

$\mathfrak{\huge{\green{\underline{Correct \: Question:-}}}}$

Find the area of the region bounded by the ellipse $\sf \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} =1$

$\mathfrak{\huge{\green{\underline{Given:-}}}}$

➤ Equation of the ellipse = $\sf \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} =1$

$\mathfrak{\huge{\green{\underline{To \: Find:-}}}}$

➤ The area of the region bounded by the ellipse.

$\mathfrak{\huge{\green{\underline{Solution:-}}}}$

Given,

Equation of the ellipse is $\sf \dfrac{x^{2}}{16}+\dfrac{y^{2}}{9} =1$

[Refer to the attachment for the figure]

Here,

$\sf a^{2}(=16) > b^{2}(=9)$

From the equation (1),

$\sf \dfrac{y^{2}}{9} =1-\dfrac{x^{2}}{16}=\dfrac{16-x^{2}}{16}$

$\implies \sf y^{2}=\dfrac{9}{16}(16-x^{2})$

$\implies \sf y^{2}=\dfrac{3}{4} (16-x^{2}) \: .... \: (2)$

For an arc of ellipse in first quadrant.

Ellipse (1) is symmetrical about x-axis and about y-axis.

Intersection of ellipse (1) and x-axis (y=0)

Putting y=1 in equation 1, we have

$\sf \dfrac{x^{2}}{16}=1 \implies x^{2}=16 \implies x= \pm 4$

Therefore, intersections of ellipse (1) with x-axis are (0, 4) and (0, -4)

Now,

Intersections of ellipse (1) with y-axis (x=0)

Putting x=0 in equation 1, we have

$\sf \dfrac{y^{2}}{9}=1 \implies y^{2}=9 \implies y= \pm 3$

Therefore, intersections of ellipse (1) with y-axis are (0, 3) and (0, -3)

Now,

Area of the region bounded by ellipse (1) = Total area shaded = 4 × Area 0AB of ellipse in first quadrant.

$\sf = 4 \left| \int\limits^4_0\: y \: dx \right|$

[ ∵ At end B of arc AB of ellipse; x=0 and at the end of A of arc AB; x=4 ]

$\sf = 4 \left| \int\limits^4_0 \dfrac{3}{4} \sqrt{16-x^2}} \: dx \right| = 4 \left| \int\limits^4_0 \dfrac{3}{4}\sqrt{4^{2}-x^{2}} \: dx \right |$

$\sf = 3 \bigg[\dfrac{x}{2}\sqrt{4^2}-x^{2}} + \dfrac{4^{2}}{2} \: sin^{-1} \dfrac{x}{4} \bigg]^4_0 \bigg[ \because \int \sqrt{a^{2}-x^{2}} dx=\dfrac{x}{2} \sqrt{a^{2}-x^{2}} +\dfrac{a^{2}}{2} sin^{-1}\dfrac{x}{a} \bigg]$

$\sf =3 \bigg[\dfrac{4}{2}\sqrt{16-16}+8 \: sin^{-1} \: 1-(0+8 \: sin^{-1}) \bigg] = 3 \bigg[0+ \dfrac{8 \pi}{2} \bigg]$

$\sf = 3(4 \pi) = \underline{\underline{12 \pi \: sq. \: units}}$

$\mathfrak{\huge{\green{\underline{Points \: to \: Note:-}}}}$

➤ The average value of a function can be calculated using integration.

➤  The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) is given by the formula: $\sf Area \: \int\limits^a_b \: y \: dx= \int\limits^a_b f(x) \: dx$

➤ The area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b is given by the formula: $\sf Area=\int\limits^a_b ; where\: f(x) \geq \: g(x) \: in \: [a, \: b]$