Question

Find the area of the region bounded by the curve x=at^2 square and y =2at

Answer 1

Answer:

Answer

Given equations are:

x = at2 ...... (1)

y = 2at ..... (2)

t = 1 ..... (3)

t = 2 ..... (4)

Equation (1) and (2) represents the parametric equation of the parabola.

Eliminating the parameter t, we get

This represents the Cartesian equation of the parabola opening towards the positive x - axis with focus at (a,0).

A rough sketch of the circle is given below: -

28.PNG

When t = 1, x = a

When t = 2, x = 4a

We have to find the area of shaded region.

Required area

= (shaded region ABCDEF)

= 2(shaded region BCDEB)

(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)

(As x is between

and the value of y varies, here y is Cartesian equation of the parabola)

(as

)

On integrating we get,

(by applying power rule)

On applying the limits we get,

Hence the area of the region bounded by the curve x = at2, y = 2at between the ordinates corresponding t = 1 and t = 2 is equal to

square units.

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Answer 2

Given:

Curve x=at^2 square and y =2at

To Find:

Find the area of the region bounded by the curve x=at^2 square and y =2at.

between t = 1 and t = 2.

Solution:

By eliminating t , we can see that the pair of equation represents a parabola.

• t = y/2a
• Substituting in x ,
• x = a (y/2a)²
• y² = 4ax .

At t = 1 ,

• y = 2a

At t= 2

• y = 4a

Therefore area bounded ,

• $\int\limits^{4a}_{2a} {x} \, dy$  = $\int\limits^{4a}_{2a} {\frac{y^{2}}{4a} } \, dy$ =  ((4a)³ - (2a)³ )/3 x 4a = 14a²/3 sq units.

The area of the region bounded by the curve x=at^2 square and y =2at and t =1 and t =2 , is 14a²/3 sq units.