Question

Trying to find the volume of the solid obtained by rotating the region bounded by y = x^2+1 and y=5-x^2 about the horiz line y=-1

Answer 1

The washer method will be easiest:

\[V = \pi \int\limits_{a}^{b} [R(x)]^{2} - [r(x)]^{2} dx\]

Answer 2

R(x) is the outer radius, r(x) is the inner radius. From a graph, you'll see that 5-x^2 is above x^2+1, so it's the R(x) and x^2+1 is r(x).

a and b will be the two points where the lines intersect, so you find them by setting the two equations equal to each other, and solving to find two x-values.

Since the radii need to be measured from the axis of rotation, which here is y = -1, you'll need to add 1 to each equation before inserting them into the integral, since the distance from y = -1 to each line is 1+y.

Answer 3

ie R(x) = 6 - x^2
and r(x) = x^2 + 2

That's because the radii are measured from the axis of rotation, which is y = -1 not the x-axis. If you plot these two on a graph (note you can plot graphs on google, just by googling: x^2+2, 6-x^2) and compare it to x^2+1, 5-x^2, you'll see they're shifted up one along the y-axis.

to get a and b, let x^2+1 = 5-x^2 and rearrange to find x.

Answer 4

\[V=π\int\limits_{-\sqrt{2}}^{\sqrt{2}}[(6-x ^{2})^{2}−(x ^{2}+2)^{2}] dx\]

Answer 5

Exact answer will be \[(772\pi \sqrt{2})/3\]
or 1143.3