#### To determine

**To find:**

The volume of a solid obtained by rotating the region bounded by the given curves about y-axis.

#### Answer

V=43π2ah+h23/2

#### Explanation

**1) Concept:**

i. If x is the radius of a typical shell, then circumference=2πx and height is y

ii. By shell method, the volume of the solid by rotating the region under the curve y=f(x) about y- axis from a to b is

V= ∫ab2πxf(x)dx

where, 0≤a≤b

**2) Given:**

The region bounded by x2-y2=a2, x=a+h rotated about the y- axis (a>0, h>0).

The graph of x2-y2=a2 is a hyperbola with right and left branches.

Solving x2-y2=a2 for y,

y2=x2- a2

Taking square root on both the sides,

y= ±x2- a2

Using shell method, find the typical approximating shell with radius x.

Therefore, the circumference is 2πx and height of each shell is

x2- a2 - - x2- a2=2x2- a2

From the figure, the limits of integration is from a to a+h.

So the total volume is

V= ∫aa+h2πx[2x2- a2] dx

Use substitution u=x2-a2.

Hence,du=2xdx , i.e. xdx=du2

For x=a, u=a2-a2=0 and

for x=a+h, u=a+h2-a2=a2+2ah+h2-a2=2ah +h2

Using this in the above integral, it becomes

V= ∫02ah+h24π[u]du2

=2π ∫02ah+h2u12du

=2π 23u3202ah+h2

V=43π2ah+h232

**Conclusion:**

The volume of the solid obtained by rotating the region bounded by the given curves

V=43π2ah+h232