Respuesta :
Answer:
The correct answer is: A = 1/2 bh V = integral A(x) from [-r, r].
Step-by-step explanation:
x^2 +y^2 = r^2 A = 1/2 bh b = 2y = 2sqrt[r^2 - x^2] so that gives A = 1/2 [ 2sqrt(r^2 - x^2)]h which gives A = sqrt[r^2 - x^2]h
The volume of the circular disk is ,
[tex]V=\frac{1}{2}\pi hr^{2}[/tex]
Given,
The radius of the circular disk is [tex]r[/tex].
Let the disk lie in [tex]xy[/tex] plane.
Let the center of the disc be at [tex](0, 0)[/tex] then we have,
[tex]x^{2} +y^{2} =r^{2}[/tex]
Let the parallel cross-sections perpendicular to the base be perpendicular to [tex]x-axis[/tex].
The elementary volume will be,
[tex]hydx=h\sqrt{r^{2}-x^{2}}dx[/tex]
Then the required volume will be,
[tex]V=\int_{}^{}dV\\=\int_{-r}^{r}h\sqrt{r^{2}-x^{2}}dx\\[/tex]
Substitute ,[tex]x=rsint[/tex] then [tex]dx=rcostdt[/tex] then from the above integral,
[tex]V=2h\int_{0}^{\frac{\pi}{2}}rcost\times rcostdt\\V=hr^{2}(t+0.5sin2t)^{0}_{\frac{\pi}{2}}\\V=\frac{1}{2}\pi hr^{2}[/tex]
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