Respuesta :
Answer:
Hence, the Area is [tex]\frac{24}{5} \;\text {and the Volume of the bounded region is } \frac{24\pi}{5}[/tex]
Step-by-step explanation:
Concept :
Area of the region bounded by two curve :
[tex]\int_a^b(R^2-r^2)dx[/tex]
volume of the solid generated by revolving the region about the x-axis is :
[tex]\int_a^b\pi(R^2-r^2)dx[/tex] where [tex]R[/tex] is long radius of bounded region from x-axis and [tex]r[/tex] is short radius of bounded region from x-axis.
Given :
[tex]y=6x[/tex] and [tex]y=6x^2[/tex]
To find :
Area of the region bounded by [tex]y=6x[/tex] and [tex]y=6x^2[/tex] and volume of the solid generated by revolving the region about the x-axis.
Explanation :
[tex]\because y=6x\;\text{and}\;y=6x^2[/tex]
[tex]\therefore 6x=6x^2\\\Rightarrow x=1[/tex]
Area of the region bounded by [tex]y=6x[/tex] and [tex]y=6x^2[/tex] is :
[tex]\int_a^b(R^2-r^2)dx=\int_0^1((6x)^2-(6x^2)^2)dx\\[/tex]
[tex]\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}[/tex]
Volume of the solid generated by revolving the region about the x-axis.
[tex]\int_0^1\pi((6x)^2-(6x^2)^2)dx=\int_0^1\pi(36x^2-36x^4)dx[/tex]
[tex]\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}[/tex]
