# Volume and Average Value

• December 2nd 2008, 02:57 PM
McDiesel
Volume and Average Value

Find the volume of the solid created by rotating the region bounded by the curves below around the x-axis.

$x=1, x=9, y=0, y= \frac{9}{\sqrt{x+5}}$

So far I have my work is as follows

$V=\pi \int_1^9 \frac{9}{\sqrt{x+5}^2}dx$

$V=\pi \int_1^9 \frac{9}{x+5}dx$

The answer is $81\pi ln \frac{7}{3}$

Your setup is almost right. You need to square the whole expression, meaning the 9 as well. That's where the 81 comes from. Bring the constants out and you have $81 \pi \int_{1}^{9} \frac{1}{x+5}dx$. The integral of 1/(x+5) is ln(x+5) and then all you do is plug in your bounds. The last part uses an identity that ln(a)-ln(b) = ln(a/b)