Question

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value of a, where a > 0.

Answer 1

If the area of the region bounded by the curve $y^2 =4ax$  and the line $x= 4a$  is $\frac{256}{3}$ Sq units, then the value of  $a$ will be $2$ .

What is area of the region bounded by the curve ?

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve  $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$  

⇒ $y=\sqrt{4ax}$

$x= 4a$,

Area of the region  $=\frac{256}{3}$  Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve  $= \[ \int_{0}^{4a} y \,dx \]$ ​

 $\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

 $\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$                                            

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$  and the line $x= 4a$  is $\frac{256}{3}$ Sq units, then the value of  $a$ will be $2$ .

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Answer 2

Answer:

a=2

Step-by-step explanation:

$area=2\int\limits^a_b {y} \, dx =2\int\limits^a_b {\sqrt{4ax} } \, dx \\=2 \times 2\sqrt{a} \frac{x^{\frac{3}{2} } }{\frac{3}{2} } ~from~~b~to~a\\=\frac{8}{3}\sqrt{a} (a^{\frac{3}{2} } -b^{\frac{3}{2} } )\\=\frac{8}{3} \sqrt{a}( (4a)^{\frac{3}{2} } -0)\\=\frac{8}{3} \sqrt{a} ((4a)\sqrt{4a} -0)\\=\frac{32 a}{3} \times 2a\\=\frac{64}{3} a^2$

$\frac{64}{3} a^2=\frac{256}{3} \\a^2=\frac{256}{3} \times \frac{3}{64} =4\\a=2 (a > 0)$

the curve is a right parabola.

here b=0,and a=4a for x

we are finding the area between x-axis and x from 0 t0 4a.

curve is symmetrical about x-axis so we multiply by 2.