Answer to Question #257708 in Civil and Environmental Engineering for toto

Question

Find the volume generated under the curve y=x^2 + 1 and the x-axis between x=0 and x=3, rotated about the x-axis *

Answer 1

Solution To find the volume, we use the formula

$V=\pi \int_{a}^b y^2dx$

where y(x) is function is rotated about the x-axis , between x=a and x=b.

For our case we get y=(x+2)/2; a=0 and b=4. Therefore

$V=\pi \int_{0}^4 (\frac {x+2} {2})^2dx=\pi \frac {(x+2)^3}{12} |_0^4=\pi (18-\frac{2}{3})=\frac {52} {3}\pi$

Answer

$V=\frac {52} {3}\pi$