#### To determine

**(a)**

**To prove:**

i) Cross sectional area of the filled taco in the plane through P perpendicular to the axis of the cylinder is Ax=r 16-x2-12r2sin2r16-x2.

ii) Find the volume of the filled taco

#### Answer

i)

Ax=r 16-x2-12r2sin2r16-x2

ii)

Vx=∫-44r·16-x2-12·r2·sin216-x2r dx

#### Explanation

**1) Concept:**

i) Use the area of sector and area of triangle to get required area.

ii) Let S be a solid that lies between y=a and y=b. If the cross sectional area of S in the plane Py through y and perpendicular to the y axis is A(y), where A is a continuous function, then volume of S is

V=∫abAydy

**2) Calculation:**

Place the round flat tortilla on a xy coordinate system. So it will look like this:.

It is a circle of radius 4 centred at origin, so the equation of the circle is

x2+y2=16

y2=16-x2

y=16-x2

So distance AB=2y=2·16-x2

The diagram for the cross sectional area of tortilla wrapped around cylinder is as below: The shaded region indicates filling

Here, AB is a chord on tortilla, which also represents arc of the circle (which is cross section of cylinder).

Let θ be the central angle, and AB represents the arc of circle as shown in the figure above, whose length is 216-x2. Also, OB=OA=r.

Now the area of the shaded region can be calculated as the difference between the area of sector and area of triangle. This is because according to the given condition, the volume of food is placed on the tortilla only up to the edge.

Area =Area of sector OAB-Area of triangle OAB

Area of sector OAB=12·θ·r2

Since the triangle is an isosceles triangle with same two side r, its area is given as

Area of triangle OAB=12·r2·sinθ

So the desired cross sectional area is

Area=12·θ·r2-12·r2·sinθ

Now, the formula for the length of an arc s:

s=rθ, θ=sr

Substitute θ in the above equation:

Area=12·sr·r2-12·r2·sinsr

But, s= AB=216-x2

Area=12·216-x2r·r2-12·r2·sin216-x2r

Simplifying,

Area, Ax=r·16-x2-12·r2·sin216-x2r

Hence the given statement of area has been proved.

The volume of the filled taco can be found out by integrating the area from -4 to 4.

Vx=∫-44r·16-x2-12·r2·sin216-x2r dx

**Conclusion:**

Vx=∫-44r·16-x2-12·r2·sin216-x2r dx

#### To determine

**(b)**

**To find:**

The value of r that maximizes volume of the taco

#### Answer

≈2.2912

#### Explanation

**1) Concept:**

Define the function Vr; then graph it as y=V(r) and y=V'(r) to get the required answer.

**2) Calculation:**

Define the function V(r) as

Vr=∫-44r·16-x2-12·r2·sin216-x2r dx

Use Mathematica to graph Vr and V'r using the command ‘Plot.’

It is as below:

Therefore, the maximum volume of about 52.94 occurs when r ≈2.2912.

**Conclusion:**

The maximum volume of about 52.94 occurs when r ≈2.2912.