# Calculus volume ..

• Jun 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 >_>
• Jun 5th 2009, 03:53 AM
Scott H
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(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}