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

Hence, the Area is
`(24)/(5) \;\text {and the Volume of the bounded region is } (24\pi)/(5)`

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((x^3)/(3)-(x^5)/(5)\right)_0^1\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left((1)/(3)-(1)/(5)\right)\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36*(2)/(15)\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=(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((x^3)/(3)-(x^5)/(5)\right)_0^1\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left((1)/(3)-(1)/(5)\right)\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi*(2)/(15)\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=(24\pi)/(5)`