Which radius bounds the region inside both curves?
You will be able to: Split a common polar region where its nearer boundary changes.
Which radius bounds the region inside both curves?
Two circular sprinkler zones overlap. A point belongs to the shared zone only if it lies within both reaches on its ray.
A useful starting point: How does an outer sector minus an inner sector give area? →
Words and symbols before equations
- Common region
- Points inside both bounded regions.
- Limiting radius
- The smaller nonnegative radius on a common ray.
- Boundary switch
- An angle where the limiting curve changes.
- Piecewise integral
- Separate integrals over intervals with different formulas.
What this picture assumes
Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. r=2 sin θ and r=2 cos θ. Only their common first-quadrant region is shaded. The smaller radius changes at θ=π/4; full area remains π/2−1 as the inspection ray moves.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- θ=0.785398 rad; 2 sin θ=1.41421, 2 cos θ=1.41421. Limiting radius=1.41421. Boundary switch π/4; full common area π/2−1=0.570796 square units.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
In the first quadrant, r=2 sin θ and r=2 cos θ are nonnegative. Their circles overlap along rays 0≤θ≤π/2.
Solve 2 sin θ=2 cos θ to get θ=π/4, where r=√2 and the Cartesian point is (1,1). Both circles also meet at the pole, reached at different angles.
For 0≤θ≤π/4, sine is smaller; for π/4≤θ≤π/2, cosine is smaller. The common region uses these smaller radii, not the difference between them.
A=(1/2)∫₀^(π/4)(2 sin θ)²dθ+(1/2)∫_(π/4)^(π/2)(2 cos θ)²dθ=π/2−1. Each part has area π/4−1/2.
A worked example, step by step
Set up and evaluate the area common to r=4 sin θ and r=4 cos θ.
- The switch remains θ=π/4 in the first quadrant.
- Use (1/2)∫₀^(π/4)16 sin²θ dθ plus (1/2)∫_(π/4)^(π/2)16 cos²θ dθ.
- Doubling every radius multiplies area by 4.
- The common area is 4(π/2−1)=2π−4 square units.
Inside both uses the smaller radius. Outer-minus-inner finds a different region: inside one boundary and outside another.
Why is the smaller radius the limiting one?
Compare with an explanation
A point farther out would leave at least one of the two regions.
Predict. Change one thing. Explain.
Move the angle across π/4. Identify which radius ends the orange ray inside the shaded common region. The full common area stays fixed while the inspected ray changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
θ=0.785398 rad; 2 sin θ=1.41421, 2 cos θ=1.41421. Limiting radius=1.41421. Boundary switch π/4; full common area π/2−1=0.570796 square units.
Read the representation: Teal shows a curve; orange marks the selected point, tangent, ray or swept region described above. Dashed segments are auxiliary comparisons. Read the model conditions and units before comparing lengths.
Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. r=2 sin θ and r=2 cos θ. Only their common first-quadrant region is shaded. The smaller radius changes at θ=π/4; full area remains π/2−1 as the inspection ray moves.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the units, coordinate rates, parameter direction, tracing count or radial boundaries. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor r=2 sin θ and r=2 cos θ, identify both geometric intersections, write the split common-area integral and evaluate it.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Intersections are the pole and (1,1).
- 1 point: The nonzero switch is θ=π/4.
- 1 point: A=2∫₀^(π/4)sin²θ dθ+2∫_(π/4)^(π/2)cos²θ dθ.
- 1 point: Each part is π/4−1/2, so total area is π/2−1.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What radius bounds an intersection of radial regions?
The smaller nonnegative radius on each ray.
RECALL 2Why check the pole separately?
Different curves can reach it at different angles.
RECALL 3What changes at π/4 here?
Which curve supplies the nearer boundary.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Which radius bounds the region inside both curves?
- For a shared radial region, use the smaller nonnegative radius.
- Split at boundary switches.
- Check the pole as well as equal-radius intersections.
Remember: Inside both uses the smaller radius. Outer-minus-inner finds a different region: inside one boundary and outside another.
Conditions: Original model. Coordinates have equal visual scales; angles are radians. Curves are sampled for display, while formulas determine the readouts. r=2 sin θ and r=2 cos θ. Only their common first-quadrant region is shaded. The smaller radius changes at θ=π/4; full area remains π/2−1 as the inspection ray moves.
Refresh Kid · AP Calculus BC Unit 9 · Objectives CHA-5.D · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.9, CHA-5.D. CED effective Fall 2020 and Fall 2026 clarifications checked September 17, 2026. This unit covers BC topics 9.1–9.9. Parametric and vector motion is planar. Polar arc length, surface area and three-dimensional vector calculus are not assigned as required Unit 9 material.
Parametric slope and second derivatives retain their nonzero-denominator conditions. Speed is the magnitude of velocity; displacement and distance use different integrals. Polar coordinates allow signed radii, with distance abs(r). Polar tangents use Cartesian angular rates. Areas use squared radial boundaries with correct angular intervals, tracing counts and boundary switches; the pole is checked separately.
All focused explanations, examples, practice and models are original Refresh Kid work. OpenStax was consulted for mathematical cross-checking. Khan Academy’s destination was checked, but JavaScript lesson content was not fully readable by the research tool. Organic Chemistry Tutor video titles, creator and destinations were checked; full videos were not reviewed. No questions, diagrams or provider scripts were copied. Resources are optional; Refresh Kid is not affiliated with or endorsed by these providers.
GitHub’s 3D website collection informed optional camera and spatial inspection. The original parameter-lift model uses self-hosted Three.js with its MIT license. The teal projection is the physical circle. The vertical axis of the orange lift is time, not spatial height, and its scale is explicitly stated. No spatial arc-length calculation is made from that lift. Keyboard controls and labeled 2D alternatives remain available, without autoplay or required WebGL.
Independent teacher review and observation of students remain pending. Technical checks do not certify mathematical accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Calculus BC questions and scoring guides. The archive spans multiple units; no entire exam question is assigned here.
Learn → Explore → Practice → Review is informed by the IES learning guide; this implementation has not been evaluated with learners.
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