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Calculus Chapters 7 & 8. Chapter 7 Resources. Chapter 7 unassigned problems that would be good review for test: 181, 183-186, 194, 195, 198. Video Cross-Sectional. Table of Contents and Introduction: General Overview of Calculus Chapter 1: Review of Derivatives Chapter 1 pdf 1.1: The Power and Exponential Rules with the Chain Rule; 1.2: Trig, Trig Inverse, and Log Rules; 1.3: Local Linearity, Euler’s Method, and Approximations.

In this topic, we will learn how to find the volume of a solid object that has known cross sections.

We consider solids whose cross sections are common shapes such as triangles, squares, rectangles, trapezoids, and semicircles.

Definition: Volume of a Solid Using Integration

Let (S) be a solid and suppose that the area of the cross section in the plane perpendicular to the (x-)axis is (Aleft( x right)) for (a le x le b.)

Then the volume of the solid from (x = a) to (x = b) is given by the cross-section formula

[V = intlimits_a^b {Aleft( x right)dx}. ]

Similarly, if the cross section is perpendicular to the (y-)axis and its area is defined by the function (Aleft( y right),) then the volume of the solid from (y = c) to (y = d) is given by

[V = intlimits_c^d {Aleft( y right)dy} .]

Steps for Finding the Volume of a Solid with a Known Cross Section

1. Sketch the base of the solid and a typical cross section.
2. Express the area of the cross section (Aleft( x right)) as a function of (x.)
3. Determine the limits of integration.
4. Evaluate the definite integral [V = intlimits_a^b {Aleft( x right)dx}.]

Solved Problems

Click or tap a problem to see the solution.

Solution.

The diameter of the semicircle at a point (x) is (d=y=x.) Hence, the area of the cross section is

[A(x) = frac{{pi {d^2}}}{8} = frac{{pi {x^2}}}{8}.]

Integration yields the following result:

[{V = intlimits_0^1 {Aleft( x right)dx} }={ intlimits_0^1 {frac{{pi {x^2}}}{8}dx} }={ frac{pi }{8}intlimits_0^1 {{x^2}dx} }={ frac{pi }{8} cdot left. {frac{{{x^3}}}{3}} right|_0^1 }={ frac{pi }{{24}}}]

Solution.

Consider an arbitrary planar section perpendicular to the (z-)axis at a point (z,) where (0 lt z le 1.) The cross section is an ellipse defined by the equation

[{z = frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}},};; Rightarrow {frac{{{x^2}}}{{{{left( {asqrt z } right)}^2}}} + frac{{{y^2}}}{{{{left( {bsqrt z } right)}^2}}} = 1.}]

The area of the cross section is

[{Aleft( z right) = pi cdot left( {asqrt z } right) cdot left( {bsqrt z } right) }={ pi abz.}]

Then, by the cross-section formula,

[{V = intlimits_0^1 {Aleft( z right)dz} }={ intlimits_0^1 {pi abzdz} }={ pi abintlimits_0^1 {zdz} }={ pi ab cdot left. {frac{{{z^2}}}{2}} right|_0^1 }={ frac{{pi ab}}{2}}]

Solution.

The area of the equilateral triangle at a point (x) is given by

[Aleft( x right) = frac{{{a^2}sqrt 3 }}{4}.]

As the side (a) is equal to (1-{x^2},) then

[{Aleft( x right) = frac{{{a^2}sqrt 3 }}{4} }={ frac{{sqrt 3 }}{4}{left( {1 – {x^2}} right)^2}.}]

The parabola (y = 1 – {x^2}) intersects the (x-)axis at the points (x=-1,) (x = 1.)

Compute the volume of the solid:

[{V = intlimits_{ – 1}^1 {Aleft( x right)dx} }={ intlimits_{ – 1}^1 {frac{{sqrt 3 }}{4}{{left( {1 – {x^2}} right)}^2}dx} }={ frac{{sqrt 3 }}{4}intlimits_{ – 1}^1 {left( {1 – 2{x^2} + {x^4}} right)dx} }={ frac{{sqrt 3 }}{4}left. {left[ {x – frac{{2{x^3}}}{3} + frac{{{x^5}}}{5}} right]} right|_{ – 1}^1 }={ frac{{sqrt 3 }}{4}left[ {left( {1 – frac{2}{3} + frac{1}{5}} right) }right.}-{left.{ left( { – 1 + frac{2}{3} – frac{1}{5}} right)} right] }={ frac{{sqrt 3 }}{2}left( {1 – frac{2}{3} + frac{1}{5}} right) }={ frac{{4sqrt 3 }}{{15}}.}]

Solution.

The area of the square cross section at a point (x) is written in the form

[Aleft( x right) = {left( {a cdot frac{x}{H}} right)^2} = frac{{{a^2}{x^2}}}{{{H^2}}}.]

Hence, the volume of the pyramid is given by

[{V = intlimits_0^H {Aleft( x right)dx} }={ intlimits_0^H {frac{{{a^2}{x^2}}}{{{H^2}}}dx} }={ frac{{{a^2}}}{{{H^2}}}intlimits_0^H {{x^2}dx} }={ frac{{{a^2}}}{{{H^2}}} cdot left. {frac{{{x^3}}}{3}} right|_0^H }={ frac{{{a^2}H}}{3}}]

Solution.

An arbitrary cross section at a point (x) has the side (a) equal to

[a = 2y = 2sqrt {1 – {x^2}} .]

Hence, the area of the cross section is

[Aleft( x right) = {a^2} = 4left( {1 – {x^2}} right).]

Calculate the volume of the solid:

[{V = intlimits_{ – 1}^1 {Aleft( x right)dx} }={ intlimits_{ – 1}^1 {4left( {1 – {x^2}} right)dx} }={ 4left. {left( {x – frac{{{x^3}}}{3}} right)} right|_{ – 1}^1 }={ 4left[ {left( {1 – frac{{{1^3}}}{3}} right) – left( { – 1 – frac{{{{left( { – 1} right)}^3}}}{3}} right)} right] }={ 4left[ {frac{2}{3} – left( { – frac{2}{3}} right)} right] }={ frac{{16}}{3}.}]

Solution.

The volume of the frustum of the cone is given by the integral

[V = intlimits_0^H {Aleft( x right)dx} ,]

where ({Aleft( x right)}) is the cross-sectional area at a point (x.)

The lengths of the major and minor axes linearly change from (a, b) to (A, B,) and at the point (x) they are determined by the following expressions:

[{text{major axis:};;a + left( {A – a} right)frac{x}{H};;}kern0pt{text{minor axis:};;b + left( {B – b} right)frac{x}{H}}.]

Let’s now calculate the area of the cross section:

[{Aleft( x right) text{ = }}kern0pt{pi left( {a + left( {A – a} right)frac{x}{H}} right)left( {b + left( {B – b} right)frac{x}{H}} right) }={ pi left[ {ab + bleft( {A – a} right)frac{x}{H} }right.}+{left.{ aleft( {B – b} right)frac{x}{H} }right.}+{left.{ left( {A – a} right)left( {B – b} right){{left( {frac{x}{H}} right)}^2}} right] }={ pi left[ {ab + left( {bA + aB – 2ab} right)frac{x}{H} }right.}+{left.{ left( {AB – aB – bA + ab} right){{left( {frac{x}{H}} right)}^2}} right].}]

Then the volume is given by

[require{cancel}{V = intlimits_0^H {Aleft( x right)dx} }={ pi left[ {abH + left( {bA + ab – 2ab} right)frac{H}{2} }right.}+{left.{ left( {AB – aB – bA + ab} right)frac{H}{3}} right] }={ pi left[ {cancel{abH} + frac{{bAH}}{2} }right.}+{left.{ frac{{aBH}}{2} – cancel{abH} }right.}+{left.{ frac{{ABH}}{3} – frac{{aBH}}{3} }right.}-{left.{ frac{{bAH}}{3} + frac{{abH}}{3}} right] }={ pi left[ {frac{{bAH}}{6} + frac{{aBH}}{6} }right.}+{left.{ frac{{ABH}}{3} + frac{{abH}}{3}} right] }={ frac{{pi H}}{6}left[ {bA + aB + 2AB + 2ab} right] }={ frac{{pi H}}{6}left[ {left( {2A + a} right)B + left( {A + 2a} right)b} right].}]

Solution.

Consider an arbitrary cross section at a height (x.) This cross section is a rectangle with the sides (m) and (n.) It is easy to see that

[{m = c + left( {a – c} right)frac{x}{h},;;}kern0pt{n = frac{{bx}}{h}.}]

Then the cross-sectional area (Aleft( x right)) is written as

[{Aleft( x right) = mn }={ left( {c + left( {a – c} right)frac{x}{h}} right)frac{{bx}}{h} }={ frac{{bcx}}{h} + frac{{ab{x^2}}}{{{h^2}}} – frac{{bc{x^2}}}{{{h^2}}} }={ frac{{bcx}}{h} + frac{{bleft( {a – c} right){x^2}}}{{{h^2}}}.}]

We find the volume of the wedge by integration:

[{V = intlimits_0^h {Aleft( x right)dx} }={ intlimits_0^h {left( {frac{{bcx}}{h} + frac{{bleft( {a – c} right){x^2}}}{{{h^2}}}} right)dx} }={ left. {frac{{bc{x^2}}}{{2h}} + frac{{bleft( {a – c} right){x^3}}}{{3{h^2}}}} right|_o^h }={ frac{{bch}}{2} + frac{{bleft( {a – c} right)h}}{3} }={ frac{{bch}}{2} + frac{{abh}}{3} – frac{{bch}}{3} }={ frac{{bch}}{6} + frac{{abh}}{3} }={ frac{{bh}}{6}left( {2a + c} right).}]

Solution.

The base of the tetrahedron is an equilateral triangle. Calculate the altitude of the base (CE.) By Pythagorean theorem,

[{{{CE}^2} = {{BC}^2} -{{EB}^2},};; Rightarrow {CE = sqrt {{BC^2} – {EB^2}} }={ sqrt {{a^2} – {{left( {frac{a}{2}} right)}^2}} }={ frac{{sqrt 3 a}}{2}.}]

Hence

[CO = frac{2}{3}CE = frac{{sqrt 3 a}}{3}.]

We express the altitude of the tetrahedron (h) in terms of (a:)

[{h = DO }={ sqrt {{DC^2} – {CO^2}} }={ sqrt {{a^2} – {{left( {frac{{sqrt 3 a}}{3}} right)}^2}} }={ sqrt {frac{{2{a^2}}}{3}} }={ frac{{sqrt 2 a}}{{sqrt 3 }}.}]

The cross-sectional area (Aleft( x right)) is written in the form

[{Aleft( x right) = frac{1}{2}{m^2}sin 60^{circ} }={ frac{{sqrt 3 {m^2}}}{4},}]

where (m) is the side of the equilateral triangle in the cross section.

It follows from the similarity that

[require{cancel}{m = frac{{ax}}{h} }={ frac{{sqrt 3 cancel{a}x}}{{sqrt 2 cancel{a}}} }={ frac{{sqrt 3 x}}{{sqrt 2 }}.}]

Using integration, we find the volume:

[{V = intlimits_0^h {Aleft( x right)dx} }={ intlimits_0^h {frac{{sqrt 3 {m^2}}}{4}dx} }={ intlimits_0^h {frac{{sqrt 3 }}{4}{{left( {frac{{sqrt 3 x}}{{sqrt 2 }}} right)}^2}dx} }={ intlimits_0^h {frac{{3sqrt 3 {x^2}}}{8}dx} }={ frac{{3sqrt 3 }}{8}intlimits_0^h {{x^2}dx} }={ frac{{3sqrt 3 }}{8} cdot left. {frac{{{x^3}}}{3}} right|_0^h }={ frac{{sqrt 3 {h^3}}}{8} }={ frac{{sqrt 3 }}{8} cdot {left( {frac{{sqrt 2 a}}{{sqrt 3 }}} right)^3} }={ frac{{2cancel{sqrt 3} sqrt 2 {a^3}}}{{24cancel{sqrt 3} }} }={ frac{{sqrt 2 {a^3}}}{{12}}.}]

Solution.

A cross section of the wedge perpendicular to the (x-)axis is a right triangle (ABC.) The leg of the triangle (AB) is given by

[AB = y = sqrt {{R^2} – {x^2}} ,]

and the other leg (BC) is expressed in the form

[{BC = AB cdot tan alpha }={ AB cdot frac{H}{R} }={ frac{H}{R}sqrt {{R^2} – {x^2}} }]

Hence, the area of the cross section is written as

[{Aleft( x right) = frac{{AB cdot BC}}{2} }={ frac{H}{R}{left( {sqrt {{R^2} – {x^2}} } right)^2} }={ frac{H}{R}left( {{R^2} – {x^2}} right).}]

Integrating yields

[{V = 2intlimits_0^R {Aleft( x right)dx} }={ 2intlimits_0^R {frac{H}{{2R}}left( {{R^2} – {x^2}} right)dx} }={ frac{H}{R}intlimits_0^R {left( {{R^2} – {x^2}} right)dx} }={ frac{H}{R}left. {left( {{R^2}x – frac{{{x^3}}}{3}} right)} right|_0^R }={ frac{H}{R} cdot frac{{2{R^3}}}{3} }={ frac{{2{R^2}H}}{3}},]

so the volume of the wedge is

[frac{{frac{{2{R^2}H}}{3}}}{{pi {R^2}H}} = frac{2}{{3pi }} approx 0.212]

of the total volume of the cylinder. The result does not depend on (R) and (H!)

7.8 Volumes With Cross Sectionsap Calculus Solver

Solution.

The figure below shows (large{frac{1}{8}}normalsize) of the solid of intersection.

Consider a cross section (ABCD) perpendicular to the (x-)axis at an arbitrary point (x). Due to symmetry, the cross section is a square with sides of length

7.8 Volumes With Cross Sectionsap Calculus Calculator

[BC = AD = y = sqrt {{R^2} – {x^2}} ,]

7.8 Volumes With Cross Sectionsap Calculus 14th Edition

[AB = CD = z = sqrt {{R^2} – {x^2}}. ]

The cross-sectional area is expressed in terms of (x) as follows:

7.8 Volumes With Cross Sectionsap Calculus 2nd Edition

[Aleft( x right) = {left( {sqrt {{R^2} – {x^2}} } right)^2} = {R^2} – {x^2}.]

Then the volume of the solid common to both the cylinders (bicylinder) is given by

[{V = 8intlimits_0^R {Aleft( x right)dx} }={ 8intlimits_0^R {left( {{R^2} – {x^2}} right)dx} }={ 8left. {left( {{R^2}x – frac{{{x^3}}}{3}} right)} right|_0^R }={ 8 cdot frac{{2{R^3}}}{3} }={ frac{{16{R^3}}}{3}}]