Question

12, find the area bounded the region by 9x = y2 and x = 4, in the first quadrant.
Please answer the 12th question only

Answer 1

Answer:

I hope it helps

Answer 2

Topic :-

Area under Curve

To Find :-

The area of region bounded by the curve 9x = y² and x = 4, in the first quadrant.

Solution :-

The given curve represented by the equation 9x = y² is a parabola symmetrical about x-axis.

The given curve represented by the equation x = 4 is a line parallel to y-axis.

We will calculate its area by taking a vertical strip of length 'dx'.

So,

Area under a curve while using a vertical strip =

$\displaystyle \int_{a}^{b}ydx$

Here,

• a = 0
• b = 4
• y² = 9x
• y = 3√x

We are taking positive value of the 'y' as we have to find the area in first quadrant.

Putting the values in formula,

$\displaystyle \int_{0}^{4}3\sqrt{x}\:dx$

$3\displaystyle \int_{0}^{4}\sqrt{x}\:dx$

$3\left [\dfrac{2x^{3/2}}{3} \right ]_0^4$

$3\left [\dfrac{2(4)^{3/2}}{3}-\dfrac{2(0)^{3/2}}{3} \right ]$

$3\left [\dfrac{2(2^2)^{3/2}}{3}-0 \right ]$

$3\left [\dfrac{2(2)^{3}}{3} \right ]$

$3\left [\dfrac{16}{3} \right ]$

$16$

Answer :-

So, the area of region bounded by the curve 9x = y² and x = 4, in the first quadrant is 16 square units.