Answer

The region *R* looks like this:

(a) With vertical slices we have to split up the region into three pieces!

The graph *y* = *x*^{4} + 1 intersects the line *y* = 5 when . The graph *y* = *x*^{2} + 1 intersects the line *y* = 5 when *x* = 2.

1) For 0 < *x* < 1 the graph of *x*^{2} + 1 is on top.

2) For the graph of *y* = *x*^{4} + 1 is on top.

3) For the graph of *y* = 5 is on top.

This means there are three pieces whose areas we need to find.

Let's look at the left piece, where 0 ≤ *x* ≤ 1.

The height of a slice at position *x* is

(*x*^{2} + 1) – (*x*^{4} + 1) = *x*^{2} – *x*^{4}.

The area of this slice is

(*x*^{2} – *x*^{4}) Δ *x*.

The area of the left piece of *R* is

Now for the middle piece, where .

The slice at position *x* has height

(*x*^{4} + 1) – (*x*^{2} + 1) = *x*^{4} – *x*^{2}.

This means the area of the slice is

(*x*^{4} – *x*^{2}) Δ *x*.

The area of the entire middle piece of *R* is

Finally, the right piece, where .

The slice at position *x* has height

5 – (*x*^{2} + 1) = 4 – *x*^{2}.

The area of this slice is

(4 – *x*^{2}) Δ *x*,

and the area of the right piece of *R* is

Putting all the integrals together, the area of *R* is

\item With horizontal slices, we need two integrals. The graphs *y* = *x*^{4} + 1 and *y* = *x*^{2} + 1 cross over each other at the point (1,2). So we need to split *R* into a lower region where 1 ≤ *y* ≤ 2 and an upper region where 2 ≤ *y* ≤ 5.

In the lower region, the length of the slice at height *y* is

(*y* – 1)^{1/4} – (*y* – 1)^{1/2}.

The area of the lower region is

In the upper region, the length of the slice at height *y* is

(*y* – 1)^{1/2} – (*y* – 1)^{1/4}.

The area of the upper region is

Adding these integrals, we find that the area of *R* is