
Calculus volume ..
graph the area bounded by y= 1 x=4 and y = radical x
fidn the volume of the solid formed by revolving the area about the line y=1
find H.
use disc/ washer method
find all teh radiuses..
my finals is in about 7 hours and im cramming alot.. and if anyone is willing to help im willing to thank them to unimaginable heights! i need to pass calc hopefully..
this will make or break my grade >_>

To find the area enclosed by
$\displaystyle \begin{aligned}
y&=1\\
x&=4\\
y&=\sqrt{x},
\end{aligned}$
we note that $\displaystyle \sqrt{x}$ intersects $\displaystyle y=1$ at $\displaystyle x=1$ and attains $\displaystyle y=2$ at $\displaystyle x=4$. Thus, the region bounded is in the first quadrant with limits $\displaystyle x=1$ and $\displaystyle x=4$. Subtracting the area under $\displaystyle y=1$ from the area under $\displaystyle y=\sqrt{x}$, we obtain
$\displaystyle \begin{aligned}
\int_1^4 (\sqrt{x}1)\,dx&=\left(\frac{2}{3}x^{\frac{3}{2}}x\right)_1^4\\
&=\left(\frac{16}{3}4\right)\left(\frac{2}{3}1\right)\\
&=\frac{5}{3}.
\end{aligned}$
To find the area of revolution, we integrate not along rectangles but along discs:
$\displaystyle \begin{aligned}
V&=\int_1^4 \pi(\sqrt{x}  1)^2\,dx\\
&=\pi\int_1^4(x2\sqrt{x}+1)\,dx\\
&=\pi\left(\frac{x^2}{2}\right)_1^42\pi\left(\frac{2}{3}x^{\frac{3}{2}}\right)_1^4+\p i(x)_1^4\\
&=\frac{15\pi}{2}\frac{28\pi}{3}+3\pi\\
&=\frac{7\pi}{6}.
\end{aligned}$