The solid whose base is the triangle with vertices ​, ​, and and whose cross sections perpendicular to the base and parallel to the​ y-axis are semicircles.

Respuesta :

The volume of the solid is [tex]\frac{a^{3}}{3}[/tex] cubic units.

How to find the volume of a solid by slice integration method

First, we need to determine the coefficients linear function behind the hypotenuse of the triangle by solving the following system of linear equations:

[tex]a\cdot m + b = 0[/tex] (1)

[tex]b = a[/tex] (2)

Where:

  • [tex]m[/tex] - Slope
  • [tex]b[/tex] - Intercept

The solution of the system is: [tex]m = -1[/tex], [tex]b = a[/tex]. Thus, the equation of the line is [tex]y = -x + a[/tex].

The integral expression for the solid volume is described below:

[tex]V = \int\limits_{0}^{a} {A(x)} \, dx[/tex] (3)

Where [tex]A(x)[/tex] is the cross section area function.

If we kwow that [tex]A(x) = \pi\cdot y^{2}[/tex], then the volume of the solid is:

[tex]V = \int\limits^a_0 {(-x+a)^{2}} \, dx[/tex] (4)

[tex]V =\int\limits^a_0 {(x^{2}-2\cdot a\cdot x + a^{2})} \, dx[/tex]

[tex]V = \int\limits^a_0 {x^{2}} \, dx -2\cdot a\int\limits^a_0 {x} \, dx + a^{2}\int\limits^a_0 {dx}[/tex]

[tex]V = \frac{a^{3}}{3}-a^{3}+a^{3}[/tex]

[tex]V = \frac{a^{3}}{3}[/tex]

The volume of the solid is [tex]\frac{a^{3}}{3}[/tex] cubic units. [tex]\blacksquare[/tex]

Remark

The statement is incomplete and poorly formatted, correct form is shown below:

Find the solid whose base is the triangle with vertices [tex](0, 0)[/tex], [tex](0, a)[/tex] and [tex](a, 0)[/tex] and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles.

To learn more on volumes, we kindly invite to check this verified question: https://brainly.com/question/1578538

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