Question

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

Answer 1

Answer:

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

Step-by-step explanation:

$dv=(Area) \ thickness$

    $=\pi r^2 \ dx\\\\=\pi (\frac{r}{h} x)^2 \ dx$  

$V= \int^{h}_{0} \pi (\frac{r}{h} x)^2 \ dx$

   $=\pi \int^{h}_{0}(\frac{r}{h} x)^2 \ dx$