Respuesta :
Answer:
Volume = 16 unit^3
Step-by-step explanation:
Given:
- Solid lies between planes x = 0 and x = 4.
- The diagonals rum from curves y = sqrt(x) to y = -sqrt(x)
Find:
Determine the Volume bounded.
Solution:
- First we will find the projected area of the solid on the x = 0 plane.
A(x) = 0.5*(diagonal)^2
- Since the diagonal run from y = sqrt(x) to y = -sqrt(x). We have,
A(x) = 0.5*(sqrt(x) + sqrt(x) )^2
A(x) = 0.5*(4x) = 2x
- Using the Area we will integrate int the direction of x from 0 to 4 too get the volume of the solid:
V = integral(A(x)).dx
V = integral(2*x).dx
V = x^2
- Evaluate limits 0 < x < 4:
V= 16 - 0 = 16 unit^3
The volume of the solid that lies between planes perpendicular to the x-axis at x=0 and x=4 is 16 cubic units.
Given
Solid lies between planes x = 0 and x = 4.
The diagonals rum from curves;
[tex]\rm y=\sqrt{x} \ to \ \sqrt{-x}[/tex]
Integration;
Integration is the reverse of differentiation.
The projected area of the solid on the x = 0 planes is;
[tex]\rm Area =0.5 \times Diagonal ^2\\\\Area =0.5 \times (\sqrt{x} +\sqrt{-x} )^2\\\\Area=0.5 \times (4x)\\\\Area =2x[/tex]
Therefore,
The volume of the solid that lies between planes perpendicular to the x-axis at x=0 and x=4 is;
[tex]\rm Volume =\int\limits^4_0 2x dx\\\\Volume =2\int\limits^4_0x dx\\\\ Volume = 2\left [\dfrac{x^2}{2} \right ]^4_0\\\\Volume =2\left [\dfrac{4^2}{2} -\dfrac{0^0}{2} \right ]\\\\Volume =22\left [\dfrac{16}{2}-0 \right ]\\\\Volume =2\left [8-0 \right ]\\\\Volume = 2\times 8\\\\Volume =16[/tex]
Hence, the volume of the solid that lies between planes perpendicular to the x-axis at x=0 and x=4 is 16 cubic units.
To know more about integration click the link given below.
https://brainly.com/question/14406733
