Answered

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Let R be the triangular region in the first quadrant with vertices at.Points (0,0), (h,0), and (h,r), where r and h are positive constants

Sagot :

Answer:

The answer is "[tex]\pi \int^{h}_{0}(\frac{r}{h} x)^2 \ dx[/tex]"

Step-by-step explanation:

[tex]dv=(Area) \ thickness[/tex]

    [tex]=\pi r^2 \ dx\\\\=\pi (\frac{r}{h} x)^2 \ dx[/tex]  

[tex]V= \int^{h}_{0} \pi (\frac{r}{h} x)^2 \ dx\\\\[/tex]

   [tex]=\pi \int^{h}_{0}(\frac{r}{h} x)^2 \ dx[/tex]

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