Unit 1 · Lesson 3 of 7

Turn a radical into a trigonometric identity.

A circular cross-section produces square roots such as sqrt( $a^2-x^2$ ). An angle parameter makes that geometry simpler.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its definite integral.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase upper x bound b? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

Read the model

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Circle height

Upper x bound b1

definite integral

I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322

The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

Substitute $x=a$ sinθ and include $dx=a$ cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Keep the reasoning visible

\sqrt{a^2-a^2\sin^2\theta}=a|\cos\theta|

The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x.

What makes this model useful?

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The absolute value is removed only on a branch where cosine is nonnegative. Different radical forms require different substitutions.

A worked example

Calculate the starting scenario

Use upper x bound b = 1 .
$I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322$
The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x. The checked quantity is definite integral.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the definite integral using upper x bound b 1 . State the meaning of its sign and units.

Checked quantity: definite integral · square units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Substitute x=a sinθ and include dx=a cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

Explore the visual solution

Read the model

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Circle height

Upper x bound b1

definite integral

I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322

The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Calculate and explain

Apply the idea to a new scenario

Integrate sqrt( $4-x^2$ ) from 0 to 1.5.

Checked quantity: definite integral · square units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

Read the model

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Circle height

Upper x bound b1.5

definite integral

I=\frac{1.5}2\sqrt{4-1.5^2}+2\arcsin(1.5/2)=2.68828

The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

Read the model

x=2\sin\theta,\quad\int_0^b\sqrt{4-x^2}\,dx=\frac b2\sqrt{4-b^2}+2\arcsin(b/2)

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

Circle height

Upper x bound b1

definite integral

I=\frac{1}2\sqrt{4-1^2}+2\arcsin(1/2)=1.91322

The substitution removes the radical by an identity; $dx=2$ cosθ dθ also changes. A reference triangle converts back to x.

Assumptions, units, and limits

Choose −π/2≤θ≤π/2 so cosθ is nonnegative. The square root is the upper circle of radius 2.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Substitute $x=a$ sinθ and include $dx=a$ cosθ dθ. Restrict the angle so the square root's sign is controlled. After integration, return to x with an inverse angle or reference triangle; do not leave a mixed-variable result.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.