Please help me solve the following problem.

"Whatr is the volume of the solid formed by revolving about the x axis the region in the first quadrant of the xy plane bounded by co ordinate axes and the graph of the equation y=(1+x^2)^(-.5)?

## GR0568 #16

### Re: GR0568 #16

For volumes of solids of revolution, you want to create a new function corresponding to the cross-sectional area of the object, and integrate over that. Your function $$f(x)$$ corresponds to the radius, so your integral will be

$$\pi \int (f(x))^2 dx$$

$$\pi \int (f(x))^2 dx$$

### Re: GR0568 #16

Volume = Int pi * y^2 dx | (inf, -inf)

with that ans should be pi^2

with that ans should be pi^2

### Re: GR0568 #16

I know rite?!, but that ain t the answer. it ((pi)^2)/2!!

### Re: GR0568 #16

Only the part in the first quadrant is revolved. Use

$$A=\pi \int_0^\infty \left[f(x)\right] ^2dx.$$

$$A=\pi \int_0^\infty \left[f(x)\right] ^2dx.$$

### Re: GR0568 #16

Thanks joey.

the limit should be 0, Inf

with that ans is pi^2/2

the limit should be 0, Inf

with that ans is pi^2/2