# Calculus volume ..

• June 4th 2009, 11:10 PM
CalculusDUMMIE
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

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 >_>
• June 5th 2009, 03:53 AM
Scott H
To find the area enclosed by

\begin{aligned}
y&=1\\
x&=4\\
y&=\sqrt{x},
\end{aligned}

we note that $\sqrt{x}$ intersects $y=1$ at $x=1$ and attains $y=2$ at $x=4$. Thus, the region bounded is in the first quadrant with limits $x=1$ and $x=4$. Subtracting the area under $y=1$ from the area under $y=\sqrt{x}$, we obtain

\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:

\begin{aligned}
V&=\int_1^4 \pi(\sqrt{x} - 1)^2\,dx\\
&=\pi\int_1^4(x-2\sqrt{x}+1)\,dx\\
&=\pi\left(\frac{x^2}{2}\right)_1^4-2\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}