#### To determine

**(a)**

A way to “slice” the water into parallel rectangular cross-sections and thenset upa definite integral for the volume of the water in the glass

#### Answer

V=∫-rrLrr-y·r2-y2dy

#### Explanation

**Concept:**

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

**Calculation:**

Consider the diagram below.

In order to take rectangular cross sections, slice the water in xz plane.

Viewing from FV forms the circle of radius r in xy plane, which can be considered as the width of rectangle, and viewing from SV forms the triangle, which can be considered as the length of the rectangle.

Equation of the circle passing through the origin:

x2+y2=r2

x2=r2-y2

x=r2-y2

Now width of rectangle is 2x.

w=2x=2r2-y2

To calculate length BC, consider ∆PQR and ∆PBC.

Let BC=z

As these triangles are similar triangles, by using similar triangle theorem,

BCQR=PBPQ

zL=r-y2r

z=L2r(r-y)

Area of rectangle is

A=L2rr-y·2r2-y2

Thus, integrating volume form -r to r

V=∫-rrL2rr-y·2r2-y2dy

V=∫-rrLrr-y·r2-y2dy

**Conclusion:**

V=∫-rrLrr-y·r2-y2dy

#### To determine

**(b)**

A way to slice water into parallel cross-sections and then set up a definite integral for the volume of the water in the glass

#### Answer

V=∫-rrLr2-x2dx

#### Explanation

**Concept:**

i) 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

ii) Integrals of Symmetric Functions:

Suppose f is continuous on [-a, a]

a) If f is even [f-x=fx], then ∫-aaf(x)dx=2∫0af(x)dx

b) If f is odd [f-x=-fx], then ∫-aaf(x)dx=0

**Calculation:**

Consider the diagram below

In order to take the trapezoid’s cross section, slice the water in yx plane.

STUV is the trapezoidal with base ST and VU and height SV.

To find height SV, consider the following diagram.

Equation of the circle passing through the origin:

x2+y2=r2

y2=r2-x2

y=r2-x2

VS=2y=2r2-x2

PA=2r

To find the length of the base, consider the following diagram:

Where, PAB is a triangle and S'T'U'V' is the projected trapezoidal projected on ∆PAB

To find S'P and V'A

S'P = V'A=12AP-SV=r-r2-x2

By similar triangles, ∆PAB and ∆PS'T',

STPS'=ABAP

ST=L2r(r-r2-x2)

By similar triangles, ∆PAB and ∆PV'U',

VUV'P=ABAP

To find V'P,

V'P=PA-V'A=2r-r-r2-x2=r+r2-x2

VU=L2r(r+r2-x2)

So, the area of the trapezoidal is

A=12(ST+VU)(SV)

A=12L2r(r-r2-x2)+L2r(r+r2-x2)(2r2-x2)

A=Lr2-x2

Thus, integrating volume form -r to r,

V=∫-rrLr2-x2dx

**Conclusion:**

V=∫-rrLr2-x2dx

#### To determine

**(c)**

**To find:**

Volume of water

#### Answer

For part a:

V=Lπr22

For part b:

V=Lπr22

#### Explanation

**Concept:**

Integrals of Symmetric Functions:

Suppose f is continuous on [-a, a].

a) If f is even [f-x=fx], then ∫-aaf(x)dx=2∫0af(x)dx

b) If f is odd [f-x=-fx], then ∫-aaf(x)dx=0

**Calculation:**

From part a,

V=L∫-rr1rr-y·r2-y2dy

V=L∫-rr1-yr·r2-y2dy

V=L∫-rrr2-y2-yrr2-y2dy

V=L∫-rrr2-y2 dy-1r∫-rryr2-y2 dy

By applying integrals of symmetric functions for second integral (it is odd, so its integral is zero),

V=L∫-rrr2-y2-0dy

V=L∫-rrr2-y2dy

Now, the above integrand is the area of the semi-circle of radius r ranging from -r to r.

Asemicircle=πr22

∫-rrr2-y2dy=πr22

V=L·πr22

V=Lπr22

From part b,

V=∫-rrLr2-x2dx

V=L∫-rrr2-x2dx

Now, the above integrand is the area of the semi-circle of radius r ranging from -r to r.

Asemicircle=πr22

∫-rrr2-y2dy=πr22

V=L·πr22

V=Lπr22

**Conclusion:**

For part a:

V=Lπr22

For part b:

V=Lπr22

#### To determine

**(d)**

**To find:**

Volume of water

#### Answer

V=12πr2L

#### Explanation

**Formula:**

Volume of cylinder V=πr2L

**Calculation:**

From the figure, the water is exactly half of the volume of the glass as the bottom filled portion of the glass is exactly equal to the top empty portion of the glass.

So, the volume of water is half the volume of the cylinder.

Vwater=12Vcylinder=12πr2L

**Conclusion:**

Vwater=12πr2L

#### To determine

**(e)**

**To find:**

Volume of water

#### Answer

23r2L

#### Explanation

**Concept:**

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

S=∫abAxdx

ii) Integrals of Symmetric Functions:

Suppose f is continuous on [-a, a].

a) If f is even [f-x=fx], then ∫-aaf(x)dx=2∫0af(x)dx

b) If f is odd [f-x=-fx], then ∫-aaf(x)dx=0

**Calculation:**

Consider the diagram below.

Draw a plane perpendicular to x axis to get a triangular section. The plane perpendicular to y axis is the rectangular cross section and the plane perpendicular to z axis is the circular cross section.

Viewing from FV forms the circle of radius r in xy plane, which can be considered as the height of triangle and viewing from SV forms the triangle, which can be considered as the length of the triangle.

Equation of the circle passing through the origin

x2+y2=r2

x2=r2-y2

y=r2-x2

AB=r2-x2

From the second diagram, as these triangles are similar triangle, by using similar triangle theorem,

BCDE=ABAD

zL=ABr

z=Lrr2-x2

Therefore, the area of the cross section is

A=12bh

A=12Lrr2-x2r2-x2

A=L2rr2-x2

Thus, integrating volume form -r to r,

V=∫-rrL2rr2-x2dx

By integrals of symmetric function,

V=2∫0rL2rr2-x2dx

V=Lr∫0rr2-x2dx

V=Lrr2x-x330r

V=Lrr3-r33

V=Lr2r33

V=2Lr23

**Conclusion:**

The volume of water is

V=2Lr23