Set up volumes, parametric lengths and surfaces of revolution

Begin with the actual region or curve

Senior 5A/5B supply integration techniques; 7D supplies parametric derivatives and speed. A solid volume, a curve length and a swept surface have different local elements. Identify the axis, interval, radii and whether each point or surface patch is covered once before calculating

Build volume from disk slices

For a region between y=f(x) and the x-axis on a≤x≤b, a perpendicular slice becomes a disk of radius |f(x)| when revolved about the x-axis. Sum its area πf(x)² times thickness dx. Coordinates can be negative; radius and volume cannot

Work a negative-side generating curve

The radius is 2x² when the region is revolved about the x-axis and coordinates are metres. The planar area ∫|y|dx is a different quantity

For the region between y=-2x² and the x-axis on 0≤x≤2, revolved about the x-axis, give V/π

Use horizontal slices about the y-axis

For x=g(y), c≤y≤d, revolving the region between that curve and the y-axis gives V=π∫cᵈg(y)²dy. The radial coordinate is x and slice thickness is dy. For x=3 y, 0≤y≤2, V=π∫₀²9 y²dy=24π

Give V/π when the region between x=3y and the y-axis on 0≤y≤2 is revolved about the y-axis

Subtract disk areas for a washer

When a cross-section leaves a hole, use its outer radius R and inner radius r, with R≥r≥0. The cross-sectional area is π(R²-r²), not π(R-r)². On a negative-side region, identify which coordinate has larger absolute distance

At x=1 in the diagram, what is the annular slice area?

Work the whole washer solid

Between y=x+2 and y=x for 0≤x≤2, rotate about the x-axis. V=π∫₀²[(x+2)²-x²]dx=π∫₀²(4 x+4)dx=16π. Rotating the corresponding negative-side region between y=-x-2 and y=-x gives the same solid

Give V/π when the region between y=x and y=x+2 on 0≤x≤2 is revolved about the x-axis

A region crossing the axis does not leave a hole

If a vertical slice is the whole interval from y=-1 to y=2, its rotation is a disk of radius 2, not an annulus. Over 0≤x≤3 this gives 12π m³. In general a slice crossing the axis has radius max(|lower|,|upper|); split where that maximum changes. Disjoint radial intervals may require unions of annuli

Give V/π when the region −1≤y≤2, 0≤x≤3 is revolved about the x-axis

Include the parameter Jacobian in volume

For x=x(t), y=y(t) on one monotone axial segment, disk volume about x uses π∫y(t)²|x′(t)|dt with increasing parameter limits. About y use π∫x(t)²|y′(t)|dt. Reversing traversal changes a signed integral, not geometric volume. If the boundary folds, split and identify the actual cross-sections instead of summing overlapping volumes

Work a parametric solid

For x=t², y=2 t, 0≤t≤3, rotate the region bounded by this curve, the x-axis and x=9 about x. Then V=π∫₀³(2 t)²(2 t)dt=162π. Eliminating t gives y²=4 x, so π∫₀⁹4 x dx confirms the answer. Traversing the same boundary from 3 down to 0 leaves volume unchanged

For x=t², y=2t on 0≤t≤3, revolve the region between the curve and the x-axis about the x-axis. Give the coefficient of t³ in the positive-volume integrand before multiplying by π

Differentiate a vector component by component

For a differentiable position vector r(t), velocity is v(t)=r′(t) and acceleration is a(t)=r″(t). In orthonormal coordinates, speed is |v|=√(vx²+vy²+vz²), omitting the third component in a planar problem. Signed velocity components are not speeds

The vector integral gives displacement

For a continuous velocity,∫ from a to b of v(t)dt=r(b)-r(a). Integrating speed instead gives distance travelled:∫ from a to b of|v(t)|dt for a<b. The norm of net displacement need not equal distance if the path curves or reverses. This distinction supplies the arc-length setup used in 5D

Derive the parametric arc-length element

A small path increment has length approximately √((dx)²+(dy)²). Factoring dt and taking the limit gives ds=√(x′²+y′²)dt for increasing t. Integrate over the complete allowed interval; in three dimensions add z′². Zero coordinate derivatives need not make the speed zero

Work a curve with a turning axial coordinate

Let x=t-t³/3, y=t², 0≤t≤2. Speed is √((1-t²)²+4 t²)=1+t², so L=[t+t³/3]₀²=14/3. The x-coordinate turns at t=1, but the path continues and the arc-length formula needs no division by x′. It does not retrace because y increases

Find the length of x=t−t³/3, y=t² on 0≤t≤2.

Circular speed and angular change

For x=h+Rcos(ωt), y=k+Rsin(ωt), speed is R|ω|. On a once-traced interval, length is radius times angular change. For R=3,ω=2, 0≤t≤π/4, speed 6 and length 3π/2. Translation does not change these values

For a circular arc of radius 3 with angular speed 2 on 0≤t≤π/4, give L/π

Surface area sums circumference times slant length

A small meridian segment ds sweeps a narrow band with area approximately 2πr ds. Thus about x use radius|y(t)| and about y use|x(t)|, each multiplied by the full speed. For y=f(x), ds=√(1+f′(x)²)dx. The usual formula gives the swept lateral surface; add planar caps separately only when requested

Work a frustum lateral surface

Rotate y=1+3 x/4, 0≤x≤4, about x. The slant factor is 5/4, so A=2π(5/4)∫₀⁴(1+3 x/4)dx=25π. The end radii 1 and 4 give cap areas π and 16π; the full closed boundary would have area 42π. The lateral calculation alone is not 42π

Give lateral A/π when y=1+3x/4 on 0≤x≤4 is revolved about the x-axis

For the frustum generated by revolving y=1+3x/4 on 0≤x≤4 about the x-axis, give the full closed-boundary area divided by π, including both circular caps

A semicircle generates a whole sphere

For x=Rcos t, y=Rsin t, 0≤t≤π, radius to the x-axis is Rsin t and speed R. Surface area is 2πR²∫₀^πsin t dt=4πR². A restricted interval[α,β] inside[0,π] gives a zone with area 2πR²(cosα-cosβ), excluding any caps

Does the displayed upper semicircle generate half or all of the sphere surface about the x-axis?

For x=3cos t, y=3sin t on π/3≤t≤2π/3, revolved about the x-axis, give the spherical zone area A/π

Parametric surfaces use speed through axial turns

For x=t-t³/3, y=2+t², 0≤t≤2, speed 1+t² and radius 2+t². Thus A=2π∫₀²(2+t²)(1+t²)dt=2π[2 t+t³+t⁵/5]₀²=(184/5)π. The radius is increasing, so the generated circles are distinct even after x reverses direction. This is a swept surface; do not infer a disk volume by ignoring the fold

Give A/π for the surface generated by revolving x=t−t³/3, y=2+t² on 0≤t≤2 about the x-axis

Check repeated path and repeated surface coverage separately

For a radius 2 circle traversed twice, total travelled length is 8π but distinct circumference is 4π. Revolving the full circle about x makes the upper and lower semicircles generate the same sphere; two turns therefore sweep it four times. The absolute-radius surface integral is 64π, whereas the distinct sphere area is 16π

A radius 2 circle is traversed twice and revolved about a diameter. What multiplicity divides the swept surface integral to give the distinct sphere area?

Derive the three geometric elements, ungraded

Explain using thin disks/annuli, polygonal path increments and narrow surface bands why volume, length and surface area have their stated integral forms. State the units, nonnegative radius, parameter regularity and single-coverage assumptions. Explain why planar caps are separate

Model: justify local approximations before taking limits

A perpendicular slice of thickness Δu has volume approximately π(R²-r²)Δu; summing and taking the limit gives volume. A differentiable path increment has norm approximately|r′(t)|Δt, giving arc length. A short meridian segment sweeps approximately circumference 2πr times its slant length, giving surface area. Continuous or piecewise smooth data allows limits piecewise; singular endpoints can be treated by limits when the integrals converge. Distinct geometry requires counting each region or surface patch once. Dimensions are length³, length and length² respectively. Caps are planar disks not swept by an open endpoint itself

Set up both-axis volumes and a crossing region, ungraded

For the region bounded by x=y², x=4 and y=0 in the first quadrant, find its volume about each coordinate axis. Then explain why rotating -1≤y≤2, 0≤x≤3 about x gives a disk instead of a washer

Model: choose horizontal or vertical slices from the same region

The first region has 0≤y≤2 and y²≤x≤4. About y, washers give π∫₀²(16-y⁴)dy=(128/5)π. About x, write 0≤x≤4 and 0≤y≤√x; disks give π∫₀⁴x dx=8π. The axis-crossing rectangle sweeps every radius between 0 and 2 at each x, hence π∫₀³4 dx=12π. Subtracting 1² would invent a hole that is filled by the interval through the axis

Justify a parameter change and repeated coverage, ungraded

For the curve x=t-t³/3, y=t², 0≤t≤2, justify its length and show that t=2-2 s, 0≤s≤1, preserves length. Then compare the travelled length and distinct circumference of a circular path traversed three times, and explain its sphere-surface coverage

Model: use absolute parameter speed and count distinct images

The original speed is 1+t² and length 14/3. With t=2-2 s, the new speed is 2[1+(2-2 s)²]; substitution maps the endpoints in reverse and gives the same length. For a radius R circle traversed three times, travel is 6πR and distinct circumference 2πR. Its revolution about a diameter sweeps the sphere six times because each semicircle already generates the whole sphere. The integral counts multiplicity; distinct area is 4πR²

Set up an ellipse arc length and surface area, ungraded

For x=3 cos t, y=2 sin t on 0≤t≤π/2, set up the arc length and the surface generated about x. Describe a numerical evaluation and independent checks. Do not replace an ellipse arc with a circle-radius formula

Model: preserve exact setup and verify technology results

Speed is √(9 sin²t+4 cos²t). Length is ∫₀^(π/2)√(9 sin²t+4 cos²t)dt. Surface area is 4π∫₀^(π/2)sin t √(9 sin²t+4 cos²t)dt, excluding any planar cap. Use radians and a convergent numerical quadrature, record precision and repeat with a tighter tolerance. Speed bounds 2≤speed≤3 give π≤length≤3π/2 and 8π≤surface≤12π. With u=cos t the surface integral becomes 4π∫₀¹√(9-5 u²)du, an independent setup check. A calculator output alone is not evidence that the correct curve, axis or interval was used

Construct and interpret a surface diagram, ungraded

Coordinates are in metres. Sketch a generating meridian for y=1+3 x/4, 0≤x≤4 and identify its axis, both endpoint radii and a small slant segment. Explain the difference between the lateral surface, a closed solid boundary and the enclosed volume. Include units and assumptions

Model: label the meridian before drawing the spatial interpretation

Coordinates are in metres. Mark endpoints(0, 1),(4, 4) above the x-axis, radii 1 and 4, axial length 4 and slant length 5. Rotation gives a frustum. Its lateral area is π(1+4)·5=25π m²; adding two end disks gives 42π m² total closed-boundary area. Its enclosed volume is(π·4/3)(1²+1·4+4²)=28π m³. These distinct values follow from the same meridian but measure different objects. The graph is a meridian, not a complete perspective rendering

Assessment boundary

Practice checks bounded exact integral factors, lengths, areas, volumes and coverage conclusions. Full derivation, unrestricted region decomposition, independent sketches and technology execution remain ungraded. Existing response kinds and shared Cartesian graphs are used; no new geometric renderer is required