Question

Calculate the area of the region bounded by y=6x and y=6x^2. And find the volume of the solid generated by revolving the region about the x-axis

Answer 1

Answer:

Hence, the Area is $\frac{24}{5} \;\text {and the Volume of the bounded region is } \frac{24\pi}{5}$

Step-by-step explanation:

Concept :

Area of the region bounded by two curve :

$\int_a^b(R^2-r^2)dx$

volume of the solid generated by revolving the region about the x-axis is :

 $\int_a^b\pi(R^2-r^2)dx$   where $R$ is long radius of bounded region from x-axis and $r$ is short radius of bounded region from x-axis.

Given :

$y=6x$ and $y=6x^2$

To find :

Area of the region bounded by $y=6x$ and $y=6x^2$ and volume of the solid generated by revolving the region about the x-axis.

Explanation :

$\because y=6x\;\text{and}\;y=6x^2$

$\therefore 6x=6x^2\\\Rightarrow x=1$

Area of the region bounded by $y=6x$ and $y=6x^2$ is :

     $\int_a^b(R^2-r^2)dx=\int_0^1((6x)^2-(6x^2)^2)dx$

$\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\int_0^1(36x^2-36x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\times\frac{2}{15}\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\frac{24}{5}$

Volume of the solid generated by revolving the region about the x-axis.

      $\int_0^1\pi((6x)^2-(6x^2)^2)dx=\int_0^1\pi(36x^2-36x^4)dx$

$\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\times\frac{2}{15}\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=\frac{24\pi}{5}$