Chapter 10 Solutions

Authored worked explanation for the original source problem.

Source review note: The preserved source answer is 2πb²/a − (2/3)πab². It is not the disc integral of the displayed ellipse and even combines terms with different length dimensions. The reviewed volume is 4πab²/3.

1. The source diagram gives $y=b\sqrt{a^2-x^2}/a$ on $-a\le x\le a$. Squaring gives $y^2=b^2(1-x^2/a^2)$.

   Why: The full ellipsoid runs from one end of its major axis to the other, and each perpendicular slice is a circle.

2. $V=\pi {{\int }_{-a}}^a{b}^2(1-x^2/a^2)dx=\pi b^2[x-x^3/(3a^2){{]}_{-a}}^a$.

   Why: The disc area is π times radius squared; integrate that area along the axis.

3. The upper value is $2a/3$ and the lower value is $-2a/3$, so $V=4\pi a{b}^2/3$.

   Why: Subtract both endpoint values. The result also has the expected three powers of length.

Reviewed answer: $V=4\pi a{b}^2/3$, for $a>0$ and $b>0$.

Takeaway: Squaring the radius and using the full axial interval keeps both the geometry and the units right.

Authored worked explanation for the original source problem.

1. For $0\le x\le 3/4$, $x\ge x^3$. Each vertical shell has radius $x$ and height $x-x^3$.

   Why: The explicit right cutoff is 0.75, even though the curves meet again at x=1.

2. $V=2\pi {{\int }_0}^{3/4}x(x-x^3)dx=2\pi [x^3/3-x^5/5{{]}_0}^{3/4}$.

   Why: Shell volume uses circumference times height times thickness.

3. Evaluation gives $477\pi /2560\approx 0.585367$, which rounds to $0.5854$.

   Why: Apply the specified interval rather than extending the region to its next intersection.

Reviewed answer: $V=477\pi /2560\approx 0.585367$ cubic units.

Takeaway: A stated vertical cutoff is part of the boundary, even when it is not an intersection.

Authored worked explanation for the original source problem.

Source review note: The source gives (4/3)πr³+C, a consistent indefinite primitive. For the requested sphere volume, V(0)=0 fixes C=0.

1. Use a separate shell radius $s$. A thin spherical shell has volume approaching $4\pi s^{2}ds$.

   Why: For a shell between s and s+Δs, the leading volume term is surface area times thickness; higher-order thickness terms vanish in the integral.

2. $V\left(r\right)={{\int }_0}^{r}4\pi s^{2}ds=[4\pi s^3/3{{]}_0}^r=4\pi r^3/3$.

   Why: Summing outward from zero radius fills the entire sphere.

3. Equivalently, integrating the area gives $V\left(r\right)=4\pi r^3/3+C$, and $V\left(0\right)=0$ forces $C=0$.

   Why: A physical volume is fixed by its zero-radius boundary condition, so its constant is not arbitrary.

Reviewed answer: $V\left(r\right)=4\pi r^3/3$, for $r\ge 0$.

Takeaway: An indefinite primitive becomes a geometric volume only after its boundary condition is applied.

Authored worked explanation for the original source problem.

1. Treat $r\left(x\right)=x-x^4$ as the vase’s interior radius in meters. It is nonnegative on $1/4\le x\le 1$, and the axial length is $1-1/4=3/4$ m.

   Why: Water capacity uses the interior cross section; no wall thickness is specified.

2. $V=\pi {{\int }_{1/4}}^1(x-x^4{)}^{2}dx=\pi {{\int }_{1/4}}^1(x^2-2x^5+x^8)dx$.

   Why: Each filled slice is a disc, and squaring the whole radius includes the cross term.

3. $V=\pi [x^3/3-x^6/3+x^9/9{{]}_{1/4}}^1=27783\pi /262144\approx 0.332958$ m³.

   Why: Subtract the lower endpoint; the source’s 0.333 is the result rounded to three decimal places.

Reviewed answer: $V=27783\pi /262144\approx 0.332958$ m³, or about $0.333$ m³.

Takeaway: A container’s capacity comes from its interior radius and has cubic length units.

Authored worked explanation for the original source problem.

Source review note: The source image repeats the curve-only wording without a shaded region. Its 58.4 matches 93π/5 when the region is closed to the y-axis between y=1 and y=8. Closing under the cubic over x=1..2 instead gives 62π/5. The reviewed answers state these assumptions openly.

1. Rotating just a curve describes a surface. To reproduce the source’s short answer, close the region against the y-axis and the horizontal lines $y=1$ and $y=8$.

   Why: The stated curve endpoints are (1,1) and (2,8), but the prompt does not say how the enclosed solid is filled.

2. For that interpretation, $x=y^{1/3}$ is the disc radius, so $V=\pi {{\int }_1}^8{y}^{2/3}dy=(3\pi /5)[y^{5/3}{{]}_1}^8=93\pi /5$.

   Why: Horizontal discs fill from the rotation axis to the curve.

3. If instead we close under the curve against $y=0$ and $x=1,2$, shells give $V=2\pi {{\int }_1}^{2}x·x^{3}dx=(2\pi /5)(32-1)=62\pi /5$.

   Why: This different bounded region has a hole of radius 1 and cannot be assigned the first region’s volume.

Reviewed answer: If the region is $0\le x\le y^{1/3}$ with $1\le y\le 8$, then $V=93\pi /5\approx 58.4336$ cubic units. If it is under $y=x^3$ with $1\le x\le 2$, then $V=62\pi /5\approx 38.9557$ cubic units.

Takeaway: State the filled region before choosing shells or discs; a curve by itself does not specify capacity.

Authored worked explanation for the original source problem.

Source review note: The source gives 15.2 mm³, which equals the signed cross-sectional area numerically and omits the stated 5 mm extrusion width. Its sketch shows a cavity below y=0, but the printed polynomial is positive near x=±2. A below-baseline clipped cavity gives about 76.2594 mm³; the full absolute-area region gives about 76.5187 mm³. The prompt needs a geometric clarification.

1. Let $f\left(x\right)=x^4-3x^2-3$. The drawing suggests a cavity from the baseline $y=0$ down to the curve, extruded through a constant width of 5 mm. But $f(\pm 2)=1>0$.

   Why: The polynomial rises above the baseline near the stated endpoints, so the drawn below-baseline cavity cannot have depth −f there.

2. Solve $x^4-3x^2-3=0$ by setting $u=x^2$. The positive root is $u=(3+\sqrt{21})/2$, so the cavity lies on $-\alpha \le x\le \alpha$ with $\alpha \approx 1.947123$.

   Why: Only the positive u-root produces real endpoints, and f is nonpositive between them.

3. $V=5{{\int }_{-\alpha }}^{\alpha }(3+3x^2-x^4)dx=10[3x+x^3-x^5/5{{]}_0}^{\alpha }$.

   Why: A prismatic slice has area width times depth; symmetry halves the algebra.

4. Using ${\alpha }^4=3{\alpha }^2+3$ simplifies this to $4\alpha (6+{\alpha }^2)\approx 76.2594$ mm³.

   Why: The root relation reduces the fifth-power endpoint expression.

5. Over the entire printed interval, the signed area is ${{\int }_{-2}}^2(-f(x\left)\right)dx=15.2$ mm². Multiplying by 5 gives 76 mm³, but this subtracts the pieces where f is positive.

   Why: A signed integral is not a geometric filling volume when the supposed depth changes sign.

6. If all area between the curve and baseline is intended, use $5{{\int }_{-2}}^2\left|f\right(x\left)\right|dx=2·76.2593658-76\approx 76.5187$ mm³.

   Why: The absolute-value model includes the two above-baseline end pieces, while the clipped cavity excludes them.

Reviewed answer: For the pictured cavity below $y=0$, $V=4\alpha (6+{\alpha }^2)\approx 76.2594$ mm³, where $\alpha =\sqrt{(3+\sqrt{21})/2}$. The full area-between-curves interpretation on $[-2,2]$ gives about $76.5187$ mm³.

Takeaway: Depth must stay nonnegative, and a cross-sectional area must be multiplied by the third dimension.

Authored worked explanation for the original source problem.

Source review note: The prose requests a y-axis rotation and calls the region a bounded leaf, but its 0..1 interval is incompatible with the actual leaf intersections. The source widget text also says x-axis rotation, contradicting the prose. The short answer 0.232 matches none of the stated y-axis interpretations; about 0.232332 is the y=0..1 washer integral before multiplying by π. All reviewed values above identify their own region.

1. For y-axis washers, invert the curves: $x=y^2$ from $y=\sqrt{x}$ with $y\ge 0$, and $x=e^{y-1}$ from $y=\ln x+1$.

   Why: Radii are horizontal distances from the y-axis, so we need x as a function of y.

2. The enclosed leaf starts at $y=1$ and ends at $y=\beta \approx 3.512862$, where $\beta =2\ln \beta +1$. Its second x-coordinate is ${\beta }^2\approx 12.340202$.

   Why: The curves meet at x=1 and at a second point beyond x=1; they do not enclose a leaf between x=0 and x=1.

3. On $1\le y\le \beta$, the outer radius is $y^2$ and the inner radius is $e^{y-1}$. Thus $V=\pi {{\int }_1}^{\beta }(y^4-e^{2y-2})dy=\pi [y^5/5-e^{2y-2}/2{{]}_1}^{\beta }\approx 97.8539$.

   Why: Between the intersections, y² is the larger nonnegative distance to the axis.

4. If the interval instead means $0<x\le 1$, the between-curves region extends down without bound. For $y<0$ it gives discs of radius $e^{y-1}$; for $0\le y\le 1$ it gives washers with outer radius $e^{y-1}$ and inner radius $y^2$.

   Why: Negative y cannot be inverted through y=√x, so those lower slices have no inner hole.

5. $V=\pi {{\int }_{-\infty }}^0{e}^{2y-2}dy+\pi {{\int }_0}^1(e^{2y-2}-y^4)dy=\pi [e^{-2}/2+(1-e^{-2})/2-1/5]=3\pi /10$.

   Why: Use an improper lower limit for the unbounded region; its exponentially narrowing slices still have finite volume.

6. If 0 and 1 are meant as y-bounds, keep only the second integral, obtaining $\pi \left[\right(1-e^{-2})/2-1/5]\approx 0.729894$.

   Why: Changing which coordinate owns the bounds changes the region and its volume.

Reviewed answer: For the actual enclosed leaf about the y-axis, $V\approx 97.8539$ cubic units. For the printed interval interpreted as $0<x\le 1$, $V=3\pi /10$. If the intended bounds are $0\le y\le 1$, $V=\pi \left[\right(1-e^{-2})/2-1/5]\approx 0.729894$.

Takeaway: Washer limits belong to the slicing variable, and an inverse curve retains its domain restrictions.

Authored worked explanation for the original source problem.

1. Place the cone’s tip at $y=0$, base at $y=h$, and side at $y=hx/r$ for $0\le x\le r$. The generating triangle lies above that side and below the base.

   Why: This orientation gives radius r at height h and zero radius at the tip.

2. Vertical shells have radius x and height $h-hx/r$. So $V=2\pi {{\int }_0}^{r}x(h-hx/r)dx=2\pi h[r^2/2-r^2/3]=\pi r^{2}h/3$.

   Why: A shell at radius x spans from the sloping side to the base plane.

3. For horizontal discs, the radius is $x=ry/h$. Hence $V=\pi {{\int }_0}^h(ry/h{)}^{2}dy=(\pi r^2/h^2\left)\right(h^3/3)=\pi r^{2}h/3$.

   Why: A disc is perpendicular to the cone’s axis; its radius grows linearly from the tip.

4. The disc method is the washer method with inner radius zero. Both integrals describe the same cone.

   Why: The two slice families fill the same solid, so they must agree.

Reviewed answer: $V=\pi r^{2}h/3$, for $r>0$ and $h>0$.

Takeaway: Changing slice direction changes the setup, while the region and its volume stay fixed.

Authored worked explanation for the original source problem.

Source review note: The preserved source answer is 59π²/16. Squaring the printed radius and integrating on 0..3π/4 instead gives 33π²/16+3π/2; the cross term contributes 3π/2, not another constant multiple of π².

1. The interior radius $r\left(x\right)=\sin \left(2x\right)+3/2$ is always at least 1/2. Use $V=\pi {{\int }_0}^{3\pi /4}\left(\sin \right(2x)+3/2{)}^{2}dx$.

   Why: A positive radius gives filled discs throughout this axial interval.

2. Expand the square to ${\sin }^2\left(2x\right)+3\sin \left(2x\right)+9/4$, then use ${\sin }^2\left(2x\right)=(1-\cos (4x\left)\right)/2$.

   Why: The radius square includes the cross term 3 sin(2x).

3. A primitive of the integrand without π is $11x/4-\sin \left(4x\right)/8-(3/2)\cos \left(2x\right)$.

   Why: Integrate the constant, cosine, and sine terms with their inside factors.

4. The endpoint difference is $33\pi /16+3/2$, so multiplying by π gives $33{\pi }^2/16+3\pi /2$.

   Why: At the upper endpoint cos(3π/2)=0; at zero cos0=1, leaving the extra 3/2.

Reviewed answer: $V=33{\pi }^2/16+3\pi /2\approx 25.0684$ cubic units.

Takeaway: A shifted radius creates a cross term whose definite integral may not vanish.

Authored worked explanation for the original source problem.

1. Use the same squared radius: $V=\pi {{\int }_0}^{2\pi }\left[{\sin }^2\right(2x)+3\sin (2x)+9/4]dx$.

   Why: Only the axial interval changes from problem 9.

2. Over this interval, ${{\int }_0}^{2\pi }\sin \left(2x\right)dx=0$ and ${{\int }_0}^{2\pi }{\sin }^2\left(2x\right)dx=\pi$.

   Why: The sine term spans whole periods and cancels; its square has average value 1/2.

3. $V=\pi [\pi +(9/4\left)\right(2\pi \left)\right]=11{\pi }^2/2$.

   Why: The constant radius contribution adds to the nonzero average of sine squared.

Reviewed answer: $V=11{\pi }^2/2=5.5{\pi }^2$ cubic units.

Takeaway: Over whole periods the sine cross term cancels, but its squared term does not.

Authored solution to an added practice problem.

1. A vertical strip has shell radius x and height $4-x^2$ on $0\le x\le 2$.

   Why: The region meets the y-axis and lies above y=0, so these strips rotate into shells without a hole at the center.

2. $V=2\pi {{\int }_0}^{2}x(4-x^2)dx=2\pi [2x^2-x^4/4{{]}_0}^2=8\pi$.

   Why: The shell circumference supplies the factor x before integration.

Reviewed answer: $V=8\pi$ cubic units.

Takeaway: Shell height is the region’s gap, while radius is distance to the axis.

Authored solution to an added practice problem.

1. The shell radius is x, and its height is $1/x$ on $1\le x\le 3$.

   Why: This interval stays away from the reciprocal function’s singularity.

2. $V=2\pi {{\int }_1}^{3}x(1/x)dx=2\pi {{\int }_1}^{3}1dx=4\pi$.

   Why: The height cancels the radius before integration, so no logarithmic primitive is needed.

Reviewed answer: $V=4\pi$ cubic units.

Takeaway: Simplify the complete volume integrand rather than integrating the boundary curve alone.

Authored solution to an added practice problem.

1. The axis $x=3$ is to the right of the region, so the shell radius is $3-x$ and its height is $x^2$.

   Why: A radius measures horizontal distance to the actual rotation axis, not always distance to x=0.

2. $V=2\pi {{\int }_0}^2(3-x)x^{2}dx=2\pi [x^3-x^4/4{{]}_0}^2=8\pi$.

   Why: All radii are positive and each strip produces a distinct shell.

Reviewed answer: $V=8\pi$ cubic units.

Takeaway: A shifted rotation axis changes the radius before it changes any integration.

Authored solution to an added practice problem.

1. At a height $0\le y\le 4$, the region runs from $x=0$ to $x=\sqrt{y}$. The shell radius is y and its horizontal length is $\sqrt{y}$.

   Why: Solving y=x² for nonnegative x gives the endpoint of each horizontal strip.

2. $V=2\pi {{\int }_0}^{4}y\sqrt{y}dy=2\pi \left[\right(2/5)y^{5/2}{{]}_0}^4=128\pi /5$.

   Why: Shells parallel to the x-axis use dy and a radius measured vertically.

3. A quick disc check is $\pi {{\int }_0}^2(4^2-x^4)dx=128\pi /5$.

   Why: At each x the region stretches from y=x² to y=4, giving outer radius 4 and inner radius x².

Reviewed answer: $V=128\pi /5$ cubic units.

Takeaway: Changing variables for shells means changing both the strip endpoints and the limits.

Authored solution to an added practice problem.

1. For $0\le y\le 1$, the horizontal strip goes from $x=0$ to $x=\sqrt{y}$. Its distance below $y=3$ is $3-y$.

   Why: The boundary y=x² is inverted on x≥0, and the axis lies above every strip.

2. $V=2\pi {{\int }_0}^1(3-y)\sqrt{y}dy=2\pi [2y^{3/2}-(2/5)y^{5/2}{{]}_0}^1=16\pi /5$.

   Why: Multiply circumference by horizontal length, then integrate through the full height range.

Reviewed answer: $V=16\pi /5$ cubic units.

Takeaway: Shell radius and shell length can depend on the same variable in different ways.

Authored solution to an added practice problem.

1. The strip’s radius is x and height is $1-x$. Its circumference is $2\pi x$, so the correct integral is $2\pi {{\int }_0}^{1}x(1-x)dx$.

   Why: A shell uses circumference times height; πx² is a disc area and does not belong in this shell setup.

2. $V=2\pi [x^2/2-x^3/3{{]}_0}^1=\pi /3$. The student’s integral gives π/6 instead.

   Why: The extra x changes the weighting of the shells and halves the result in this example.

Reviewed answer: $V=\pi /3$ cubic units; replace the extra radius factor.

Takeaway: Identify the slice shape before deciding whether the radius is squared.

Authored solution to an added practice problem.

1. The outer radius is $x+1$ and the inner radius is 1 on $0\le x\le 2$.

   Why: Both boundary curves lie above the x-axis, and the space below y=1 stays empty.

2. $V=\pi {{\int }_0}^2\left[\right(x+1{)}^2-1^2]dx=\pi {{\int }_0}^2(x^2+2x)dx=20\pi /3$.

   Why: Subtract the two circular areas; do not square the gap between the radii.

Reviewed answer: $V=20\pi /3$ cubic units.

Takeaway: Washer area is the difference of squares, not the square of the difference.

Authored solution to an added practice problem.

1. At height y, the region extends from the y-axis to $x=y^2$, so its disc radius is $y^2$.

   Why: A horizontal slice is perpendicular to the vertical rotation axis.

2. $V=\pi {{\int }_0}^2(y^2{)}^{2}dy=\pi [y^5/5{{]}_0}^2=32\pi /5$.

   Why: Square the complete radius before taking the primitive.

Reviewed answer: $V=32\pi /5$ cubic units.

Takeaway: The power in a boundary equation doubles when that boundary becomes a disc radius.

Authored solution to an added practice problem.

1. Distances from $y=-1$ are $R=\sqrt{x}+1$ and $r=1$ on $0\le x\le 4$.

   Why: The lower edge y=0 is one unit from the axis, so there is an inner hole.

2. $V=\pi {{\int }_0}^4\left[\right(\sqrt{x}+1{)}^2-1]dx=\pi {{\int }_0}^4(x+2\sqrt{x})dx$.

   Why: The constant squares cancel, but the cross term survives.

3. $V=\pi [x^2/2+(4/3)x^{3/2}{{]}_0}^4=56\pi /3$.

   Why: The upper endpoint contributes 8+32/3.

Reviewed answer: $V=56\pi /3$ cubic units.

Takeaway: Translate both radii to the new axis before forming the washer area.

Authored solution to an added practice problem.

1. Horizontal slices have outer radius $R=4-(-2)=6$ and inner radius $r=y^2-(-2)=y^2+2$.

   Why: The axis x=−2 lies left of both vertical boundaries, so subtracting its coordinate gives each distance.

2. $V=\pi {{\int }_0}^2[36-(y^2+2{)}^2]dy=\pi {{\int }_0}^2(32-4y^2-y^4)dy$.

   Why: For a vertical axis, washers advance in the y-direction.

3. $V=\pi [32y-(4/3)y^3-y^5/5{{]}_0}^2=704\pi /15$.

   Why: At y=2 the radii coincide, so the final washer correctly closes to zero area.

Reviewed answer: $V=704\pi /15$ cubic units.

Takeaway: The larger horizontal distance is the outer radius, regardless of which curve is algebraically simpler.

Authored solution to an added practice problem.

1. For each $-1\le x\le 1$, the region lies between y=0 and y=x. Its disc radius is $\left|x\right|$.

   Why: The boundary is below the axis when x<0, but a geometric radius is still nonnegative.

2. $V=\pi {{\int }_{-1}}^1|x{|}^{2}dx=\pi {{\int }_{-1}}^1{x}^{2}dx=2\pi /3$.

   Why: Squaring removes the radius’s sign, and the two symmetric halves add rather than cancel.

Reviewed answer: $V=2\pi /3$ cubic units.

Takeaway: A below-axis region contributes positive volume; signed area cancellation does not apply to discs.

Authored solution to an added practice problem.

1. Each vertical segment runs from y=−1 to $y=2+x$ and crosses the x-axis. Its rotation fills all radii from zero to $2+x$.

   Why: A segment crossing the axis leaves no empty central circle, and 2+x is always farther from the axis than 1.

2. Use $V=\pi {{\int }_0}^1(2+x{)}^{2}dx=\pi [(2+x{)}^3/3{{]}_0}^1=19\pi /3$.

   Why: The radius 1 describes already-filled material, so subtracting its area would remove volume that is present.

Reviewed answer: $V=19\pi /3$ cubic units; the proposed washers should be discs.

Takeaway: A washer’s inner radius comes from a real hole, not merely a second boundary curve.

Authored solution to an added practice problem.

1. At coordinate x, the sphere has disc radius $\sqrt{R^2-x^2}$. Keep only $0\le x\le R/2$.

   Why: The sphere equation determines the perpendicular circular cross section; the given planes set its axial bounds.

2. $V=\pi {{\int }_0}^{R/2}(R^2-x^2)dx=\pi [R^{2}x-x^3/3{{]}_0}^{R/2}=11\pi R^3/24$.

   Why: This is a segment of the sphere, so do not integrate over −R..R.

Reviewed answer: $V=11\pi R^3/24$, for $R>0$.

Takeaway: The same cross-sectional formula can describe different solids when its bounds change.

Authored solution to an added practice problem.

1. Let $\rho \left(x\right)=r+(R-r)x/h$ for $0\le x\le h$.

   Why: A straight conical side makes the radius interpolate linearly between the two stated end radii.

2. $V=\pi {{\int }_0}^h\rho (x{)}^{2}dx=\pi h[r^2+r(R-r)+(R-r{)}^2/3]$.

   Why: Expanding the radius square gives a constant, a linear term, and a quadratic term; integrate each over the axial length.

3. Combine the terms to get $\pi h(r^2+rR+R^2)/3$. When $r=0$ this reduces to the cone formula.

   Why: The algebra collects both squared end radii and their cross term; the zero-radius case is a useful check.

Reviewed answer: $V=\pi h(r^2+rR+R^2)/3$.

Takeaway: A radius that varies linearly still produces a quadratic cross-sectional area.

Authored solution to an added practice problem.

1. The boundaries meet at $x=0,2$, and the vertical gap is $s\left(x\right)=2x-x^2$ cm between them.

   Why: This gap is the stated side length of each square cross section and is nonnegative on the interval.

2. The cross-sectional area is $A\left(x\right)=(2x-x^2{)}^2$ cm², so $V={{\int }_0}^2(4x^2-4x^3+x^4)dx$.

   Why: The solid uses squares rather than circles; their area formula replaces πr².

3. $V=\left[\right(4/3)x^3-x^4+x^5/5{{]}_0}^2=16/15$ cm³.

   Why: Area times axial thickness has cubic centimeter units.

Reviewed answer: $V=16/15$ cm³.

Takeaway: The volume principle is still integral of area even when the slices are not circular.

Authored solution to an added practice problem.

1. At each $-2\le x\le 2$, the base disk $x^2+y^2\le 4$ has vertical gap $d\left(x\right)=2\sqrt{4-x^2}$.

   Why: The full gap is the diameter of the semicircular section, not its radius.

2. A semicircle has area $A=\pi d^2/8$, so $A\left(x\right)=\pi (4-x^2)/2$.

   Why: Use half of π(d/2)² to account for both the diameter-to-radius conversion and the half circle.

3. $V=(\pi /2){{\int }_{-2}}^2(4-x^2)dx=16\pi /3$.

   Why: Integrate the semicircular areas across the full width of the base.

Reviewed answer: $V=16\pi /3$ cubic units.

Takeaway: If a problem specifies a diameter, convert it before squaring.

Authored solution to an added practice problem.

1. The base’s vertical gap is $s\left(x\right)=\sqrt{x}$ for $0\le x\le 9$. Each triangle therefore has area $A\left(x\right)=(\sqrt{3}/4)x$.

   Why: The supplied area formula squares the side length, which removes the boundary’s square root.

2. $V=(\sqrt{3}/4){{\int }_0}^{9}xdx=(\sqrt{3}/4)(81/2)=81\sqrt{3}/8$.

   Why: Sum the triangular areas along the x-axis rather than applying a circle or shell formula.

Reviewed answer: $V=81\sqrt{3}/8$ cubic units.

Takeaway: Translate the base gap through the stated cross-sectional shape before integrating.

Authored solution to an added practice problem.

1. Split the volume integral where the cross-sectional formula changes: $V={{\int }_0}^1{x}^{2}dx+{{\int }_1}^3(4-x)dx$.

   Why: A piecewise area must be integrated with the correct formula on each axial interval.

2. The first part is $[x^3/3{{]}_0}^1=1/3$ m³; the second is $[4x-x^2/2{{]}_1}^3=4$ m³.

   Why: The second interval has length two, so its endpoint values cannot be reused from the first piece.

3. Add the two positive parts: $V=1/3+4=13/3$ m³.

   Why: The solid’s volume is the sum of its nonoverlapping axial pieces; a single jump in area does not affect the integral.

Reviewed answer: $V=13/3$ m³.

Takeaway: A change in slice formula requires a change in the integral’s pieces.

Authored solution to an added practice problem.

1. On the interval from 0 to 1 cm, the trapezoid contributes $\left(1\right)(4+7)/2=5.5$ cm³.

   Why: Average the endpoint areas and multiply by that interval’s length.

2. On the interval from 1 to 3 cm, it contributes $\left(2\right)(7+5)/2=12$ cm³.

   Why: These samples have unequal spacing; the second slice interval is twice as long.

3. Add the estimates: $5.5+12=17.5$ cm³.

   Why: The table does not specify area between samples, so trapezoids provide an estimate rather than a guaranteed exact value.

Reviewed answer: Trapezoidal estimate: $V\approx 35/2=17.5$ cm³.

Takeaway: Measured areas estimate volume only after they are weighted by their axial spacing.

Authored solution to an added practice problem.

1. The curves $y=x$ and $y=x^2$ meet at x=0 and x=1, with x above x² between them. Washers about the x-axis have radii x and x².

   Why: Both radii are nonnegative, and vertical slices are perpendicular to the axis.

2. $V=\pi {{\int }_0}^1(x^2-x^4)dx=\pi (1/3-1/5)=2\pi /15$.

   Why: Subtract the squared inner radius from the squared outer radius.

3. At height $0\le y\le 1$, a horizontal strip runs from $x=y$ to $x=\sqrt{y}$. Its shell radius is y and length is $\sqrt{y}-y$.

   Why: Invert both boundaries on their nonnegative branches before changing the slicing direction.

4. $V=2\pi {{\int }_0}^{1}y(\sqrt{y}-y)dy=2\pi (2/5-1/3)=2\pi /15$.

   Why: The shell circumference and horizontal length describe the same solid’s volume with dy.

Reviewed answer: $V=2\pi /15$ cubic units by either method.

Takeaway: Two correct setups can use different variables and still agree because their geometry agrees.