$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\[-0.5em]
\hrulefill\\
a=8\\
b=\stackrel{DC}{16}\\
h=4\sqrt{3}
\end{cases}\implies A=\cfrac{4\sqrt{3}(8+16)}{2}
\\\\\\
A=2\sqrt{3}(24)\implies \boxed{A=48\sqrt{3}}$

now, check the picture below on the bottom side.

since we know x = 9, then b = 9, therefore DC = 9+6+9 = 24, and h = b = 9.

$\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\[-0.5em]
\hrulefill\\
a=6\\
b=\stackrel{DC}{24}\\
h=9
\end{cases}\implies A=\cfrac{9(6+24)}{2}
\\\\\\
A=\cfrac{9(30)}{2}\implies \boxed{A=135}$

7 0

3 years ago

Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about th

Nezavi [6.7K]

The two curves intersect when

$-x^2=6-5x\implies x^2-5x+6=(x-3)(x-2)=0\implies x=2,x=3$

The region bounded by $x=0$ and the two curves "ends" with a vertex where $x=2$ . So over the interval [0, 2], we have $6-5x\ge-x^2$ , so that the volume is

$\displaystyle2\pi\int_0^2x(6-5x+x^2)\,\mathrm dx=\boxed{\frac{16\pi}3}$

That is, each cylindrical shell has a radius of $x$ and height $(6-5x)-(-x^2)=6-5x+x^2$ , and their contribution to the total volume is $2\pi$ times the radius and height.

5 0

3 years ago