A solid's cross-sectional area is in square feet and its height in feet. What units does the volume have?

Cubic feet. Each slice has an area in square feet and a thickness in feet, so a slice volume is in cubic feet.

Which length must you use as the height of a triangular cross section?

The perpendicular height, measured at right angles to the base. A slanting side is longer than the perpendicular height and would give too large an area.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 8: Applications of IntegrationVolumes with Cross Sections

Volumes with Cross Sections (College Board AP® Calculus BC): Flashcards

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FrontVolumes from Areas of Known Cross Sections
A solid has cross-sectional area $A (x)$ at each value of $x$ , continuous on $[a, b]$ . Its volume from $x = a$ to $x = b$ is
$\int_{a}^{b}______d x$

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Cards in this collection (29)

A solid has cross-sectional area $A (x)$ at each value of $x$ , continuous on $[a, b]$ . Its volume from $x = a$ to $x = b$ is
$\int_{a}^{b}______d x$
The completed formula is $\int_{a}^{b} A (x) d x$ .
Each slice of thickness $d x$ has volume $A (x) d x$ .
The integral adds every slice between the two limits.
What two conditions must a cross-sectional area satisfy before you can integrate it?
It must be expressible as a function of one variable, written $A (x)$ .
It must also be continuous on the interval $[a, b]$ you are integrating across.
True or False?
Finding a volume from cross-sectional areas is an application of the definite integral as an accumulation of change.
True.
$A (x) \cdot Δ x$ is the volume of a slab of thickness $Δ x$ , and $A (x) d x$ is its limit as $Δ x \to 0$ .
The integral accumulates these volume elements from $x = a$ to $x = b$ .
The cross-sectional area of a solid is $A (x) = \frac{1}{x + 1}$ for $0 \leq x \leq 3$ . Find its volume.
Substitute into the volume formula:
$\int_{0}^{3} \frac{1}{x + 1} d x = {[\ln (x + 1)]}_{0}^{3} = \ln 4 - \ln 1 = \ln 4$
A solid's cross-sectional area is in square feet and its height in feet. What units does the volume have?
Cubic feet.
Each slice has an area in square feet and a thickness in feet, so a slice volume is in cubic feet.
The horizontal cross-sectional area of a water tank at height $h$ feet is $f (h) = \frac{50}{e^{h}}$ square feet. The tank is 10 feet tall. Find its volume.
Integrate with respect to $h$ from 0 to 10.
$50 \int_{0}^{10} e^{- h} d h = 50 {[- e^{- h}]}_{0}^{10} = 50 (1 - e^{- 10}) = 49.998$ cubic feet, to 3 decimal places.
A solid is sliced perpendicular to the $y$ -axis rather than the $x$ -axis. How does the volume formula change?
The cross-sectional area becomes a function of $y$ and you integrate with respect to $y$ .
The structure is unchanged, giving $\int_{c}^{d} A (y) d y$ between the two $y$ -values.
Region $R$ lies between the graph of $f$ and the $x$ -axis. Each cross section perpendicular to the $x$ -axis is a square, so
$A (x) = {(______)}^{2}$
The completed area function is $A (x) = {(f (x))}^{2}$ .
The side of each square is the height of the region at that value of $x$ , so squaring it gives the area.
You are given the base region but no area function. What must you do?
Build the cross-sectional area function yourself.
Express the dimensions of the cross section in terms of the function or functions the question gives you, then write down $A (x)$ .
True or False?
For a solid with square cross sections you can integrate $f (x)$ first and square the result afterwards.
False.
Square first, then integrate.
${(\int f (x) d x)}^{2}$ is not the same as $\int {[f (x)]}^{2} d x$ , and only the second gives the volume.
Region $R$ is bounded by $f (x) = 1 + e^{- x}$ , the axes and $x = 3$ . Cross sections perpendicular to the $x$ -axis are squares. Find the volume.
First build the area function.
$A (x) = {(1 + e^{- x})}^{2} = 1 + 2 e^{- x} + e^{- 2 x}$ , so the volume is ${[x - 2 e^{- x} - \frac{1}{2} e^{- 2 x}]}_{0}^{3} = \frac{11}{2} - 2 e^{- 3} - \frac{1}{2} e^{- 6} = 5.399$ units cubed.
A square-based right pyramid has height $h$ and base side $2 a$ . Its sloping edge lies on $y = a - \frac{a}{h} x$ . What is $A (x)$ ?
Each square has side twice the distance from the axis up to the sloping edge.
$A (x) = {(2 (a - \frac{a}{h} x))}^{2} = 4 a^{2} {(1 - \frac{x}{h})}^{2}$
Integrating $4 a^{2} {(1 - \frac{x}{h})}^{2}$ from $0$ to $h$ gives a pyramid's volume. What is it?
It is $\frac{4}{3} a^{2} h$ .
This agrees with one third of the base area times the height, since the base area is ${(2 a)}^{2} = 4 a^{2}$ .
The base of a solid is the region between $f$ and the $x$ -axis. Each cross section perpendicular to the $x$ -axis is a rectangle of height $h (x)$ , so
$A (x) = f (x) \times______$
The completed area function is $A (x) = f (x) \times h (x)$ .
The base region supplies one side of the rectangle and the question supplies the other, so multiply the two.
Why is the base region alone not enough for a rectangular cross section?
The base fixes only one side of the rectangle.
A rectangle needs two dimensions, so the question must also give the height at each value of $x$ .
Region $R$ is bounded by $f (x) = 54 - 2 x^{3}$ , the axes and $x = 2$ . Cross sections perpendicular to the $x$ -axis are rectangles of height $h (x) = 1 + x$ . Find the volume.
Multiply the two functions to get the area.
$A (x) = (54 - 2 x^{3}) (1 + x)$ , and $\int_{0}^{2} (54 + 54 x - 2 x^{3} - 2 x^{4}) d x = \frac{976}{5} = 195.2$ units cubed.
True or False?
For rectangular cross sections you find $A (x)$ by squaring $f (x)$ .
False.
Squaring only works when both sides of the cross section equal $f (x)$ .
A rectangle has its height given separately, so $A (x) = f (x) \cdot h (x)$ .
Rectangular cross sections have width $f (x)$ and height $h (x) = 1 + x$ . If $f (2) = 38$ , find $A (2)$ .
Evaluate both functions at $x = 2$ and multiply.
$A (2) = 38 \times 3 = 114$
A solid's cross sections perpendicular to the $x$ -axis are equilateral triangles of side $f (x)$ . The cross-sectional area is
$A (x) = \frac{\sqrt{3}}{______} {[f (x)]}^{2}$
The completed area function is $A (x) = \frac{\sqrt{3}}{4} {[f (x)]}^{2}$ .
Half the base $f (x)$ times the perpendicular height $\frac{\sqrt{3}}{2} f (x)$ gives exactly that.
An equilateral triangle has side $f (x)$ . What is its perpendicular height?
It is $\frac{\sqrt{3}}{2} f (x)$ .
Cutting the triangle in half gives a right triangle with hypotenuse $f (x)$ and base $\frac{f (x)}{2}$ , so Pythagoras gives the height.
Which length must you use as the height of a triangular cross section?
The perpendicular height, measured at right angles to the base.
A slanting side is longer than the perpendicular height and would give too large an area.
True or False?
A triangular cross section always has half the area of a square cross section on the same base.
False.
That would need the triangle's perpendicular height to equal its base.
An equilateral triangle on base $f (x)$ has area $\frac{\sqrt{3}}{4} {[f (x)]}^{2}$ , about $0.433$ of the square's area.
Region $R$ is bounded by $f (x) = \sqrt{4 - x}$ and the axes, with cross sections perpendicular to the $x$ -axis that are equilateral triangles. Find the volume.
The area function is $A (x) = \frac{\sqrt{3}}{4} (4 - x)$ , since ${[f (x)]}^{2} = 4 - x$ .
Then $\frac{\sqrt{3}}{4} \int_{0}^{4} (4 - x) d x = \frac{\sqrt{3}}{4} \times 8 = 2 \sqrt{3}$ .
Why does the integral for $f (x) = \sqrt{4 - x}$ bounded by the axes run from $0$ to $4$ ?
The region begins at the $y$ -axis, where $x = 0$ , and ends where the curve meets the $x$ -axis.
Solving $\sqrt{4 - x} = 0$ gives $x = 4$ .
What is the area of a semicircle of radius $r$ ?
It is $\frac{1}{2} π r^{2}$ , half the area of the full circle with the same radius.
The base of a solid is the triangle with vertices $(0, 0)$ , $(0, 2)$ and $(4, 0)$ . What is its height at each $x$ ?
It is $2 - \frac{1}{2} x$ .
The sloping side joins $(0, 2)$ to $(4, 0)$ , so its gradient is $- \frac{1}{2}$ and its intercept is $2$ .
True or False?
For semicircular cross sections on a base of height $f (x)$ , the area is $\frac{1}{2} π {[f (x)]}^{2}$ .
False.
The height of the base is the semicircle's diameter, not its radius.
The area is $\frac{1}{2} π {(\frac{f (x)}{2})}^{2}$ , which is four times smaller.
The base region has height $2 - \frac{1}{2} x$ and each cross section perpendicular to the $x$ -axis is a semicircle. Its area is
$A (x) = \frac{1}{2} π {(______)}^{2}$
The completed area function is $A (x) = \frac{1}{2} π {(1 - \frac{1}{4} x)}^{2}$ .
The radius is $1 - \frac{1}{4} x$ , half of $2 - \frac{1}{2} x$ , since the height of the region is the diameter.
With $A (x) = \frac{π}{32} (16 - 8 x + x^{2})$ on $0 \leq x \leq 4$ , find the volume.
The volume is $\frac{π}{32} {[16 x - 4 x^{2} + \frac{1}{3} x^{3}]}_{0}^{4} = \frac{π}{32} \times \frac{64}{3} = \frac{2 π}{3} = 2.094$ units cubed, to 3 decimal places.

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