Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. y = (7x + 6)^(-1/2) y = 0 x = 0 x = 4

Answer 1

Answer:

V = .779 (that's with pi multiplied in)

Step-by-step explanation:

We will use the disk method here with a vertical representative rectangle.  With this method, we use y = x equations and x intervals.  The height of the rectangle is the curve itself, and since there is no space between the solid and the axis of revolution, the integral looks like this:

$V=\pi \int\limits^4_0 {(\frac{1}{\sqrt{7x+6} })^2-0^2 } \, dx$

When you square a square root it's eliminated, so we simply have:

$V=\pi\int\limits^4_0 {\frac{1}{7x+6} } \, dx$

Now we integrate.  This pattern follows that of the natural log.  Identify the "u" we need and take it from there.

u = 7x + 6

$\frac{du}{dx}=7$

du = 7 dx and, finally,

$\frac{1}{7}du=dx$. Our integral is as follows:

$\frac{\pi }{7}$[ln(7x+6)] from 0 to 4.

Filling in our bounds applying the First Fundamental Theorem of Calculus:

$\frac{\pi }{7}$[3.526360525 - 1.791759469] and

$\frac{1.734601\pi }{7}=.77848$