Engineering Mathematics II • Digital Workbook • Page 18

Green’s Theorem and Area

Area of an Astroid and a Generalised Hypocycloid

Green’s theorem and area are developed through three IOE past-paper questions. Use the shaded curves, positive orientation, parametric forms and symmetry.

UnitVector Calculus
TopicGreen’s Theorem
Question Type4 Mark Problems

Theory Required for These Problems

Green’s Theorem

For a positively oriented simple closed curve \(C\) enclosing a region \(R\),

\[\oint_C(F_1\,dx+F_2\,dy)=\iint_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA.\]
\[\oint_C(F_1\,dx+F_2\,dy)\]\[=\iint_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA.\]

Area Formula

Choose

\[F_1=-\frac{y}{2},\qquad F_2=\frac{x}{2}.\]

Then

\[\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=\frac12-\left(-\frac12\right)=1.\]
\[\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\]\[=\frac12-\left(-\frac12\right)=1.\]

Therefore,

\[A=\iint_Rdx\,dy=\frac12\oint_C(x\,dy-y\,dx).\]
\[A=\iint_Rdx\,dy\]\[=\frac12\oint_C(x\,dy-y\,dx).\]

The positive orientation is counterclockwise.

Parametric Form

For the astroid, use

\[x=a\cos^3t,\qquad y=a\sin^3t.\]

For the generalised form, use

\[x=a\cos^3t,\qquad y=b\sin^3t.\]

Symmetry

The curve is symmetric in all four quadrants. Therefore, calculate the first-quadrant contribution and multiply by four.

Useful result: \(\displaystyle\int_0^{\pi/2}\sin^2t\cos^2t\,dt=\frac{\pi}{16}\).

Note: Whenever problems involving a circle, ellipse, astroid, hypocycloid, etc. occur in vector integration, we prefer the parametric form.
1. Circle a x y

Cartesian form

\[x^2+y^2=a^2\]

Parametric form

\[x=a\cos t,\qquad y=a\sin t\] \[0\le t\le2\pi\]
2. Ellipse a b x y

Cartesian form

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

Parametric form

\[x=a\cos t,\qquad y=b\sin t\] \[0\le t\le2\pi\]
3. Astroid a x y

Cartesian form

\[x^{2/3}+y^{2/3}=a^{2/3}\]

Parametric form

\[x=a\cos^3t,\qquad y=a\sin^3t\] \[0\le t\le2\pi\]
4. Generalised Hypocycloid a b x y

Cartesian form

\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1\]
\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1\]

Parametric form

\[x=a\cos^3t,\qquad y=b\sin^3t\] \[0\le t\le2\pi\]
Method used: State Green’s theorem, choose \(F_1=-\frac{y}{2}\) and \(F_2=\frac{x}{2}\), write \(A=\frac12\oint_C(x\,dy-y\,dx)\), convert the Cartesian equation into parametric form, use symmetry, and evaluate the resulting trigonometric integral.
Continue through the IOE Engineering Mathematics II Digital Workbook. This green’s theorem and area page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
31

Area of an Astroid

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 10 • 4 Marks

State Green’s theorem in the plane. Apply it to find the area enclosed by

\[x^{2/3}+y^{2/3}=a^{2/3}.\]

Solution

Green’s theorem and parametrisation
Shaded region enclosed by an astroid
The astroid is symmetric about both coordinate axes.

By Green’s theorem,

\[A=\iint_Rdx\,dy=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]
\[A=\iint_Rdx\,dy\]\[=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]

The Cartesian equation of the astroid is

\[x^{2/3}+y^{2/3}=a^{2/3}.\]

Use the parametric form

\[x=a\cos^3t,\qquad y=a\sin^3t,\qquad0\le t\le2\pi.\]
\[x=a\cos^3t\]\[y=a\sin^3t\]\[0\le t\le2\pi.\]

Indeed,

\[x^{2/3}+y^{2/3}=a^{2/3}\cos^2t+a^{2/3}\sin^2t=a^{2/3}.\]
\[x^{2/3}+y^{2/3}\]\[=a^{2/3}\cos^2t+a^{2/3}\sin^2t\]\[=a^{2/3}.\]

Differentiating the parametric equations,

\[dx=-3a\cos^2t\sin t\,dt\]\[dy=3a\sin^2t\cos t\,dt.\]

Substituting the parametric values of \(x\), \(y\), \(dx\) and \(dy\),

\[A=\frac12\int_0^{2\pi}\left[(a\cos^3t)(3a\sin^2t\cos t)-(a\sin^3t)(-3a\cos^2t\sin t)\right]dt.\]
\[A=\frac12\int_0^{2\pi}\Big[(a\cos^3t)(3a\sin^2t\cos t)\]\[{}-(a\sin^3t)(-3a\cos^2t\sin t)\Big]dt.\]

Therefore,

\[A=\frac{3a^2}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t+\sin^4t\cos^2t\right)dt.\]
\[A=\frac{3a^2}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t\right.\]\[\left.{}+\sin^4t\cos^2t\right)dt.\]

Taking \(\sin^2t\cos^2t\) as the common factor,

\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\left(\cos^2t+\sin^2t\right)dt.\]
\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\]\[\qquad\cdot\left(\cos^2t+\sin^2t\right)dt.\]

Since \(\cos^2t+\sin^2t=1\),

\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\,dt.\]

The astroid is symmetric in all four quadrants. Hence,

\[A=4\left(\frac{3a^2}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]
\[A=4\left(\frac{3a^2}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]

Therefore,

\[A=6a^2\int_0^{\pi/2}\sin^2t\cos^2t\,dt.\]

Now,

\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt=\frac14\int_0^{\pi/2}\sin^22t\,dt=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt=\frac{\pi}{16}.\]
\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt\]\[=\frac14\int_0^{\pi/2}\sin^22t\,dt\]\[=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt\]\[=\frac{\pi}{16}.\]

Hence,

\[A=6a^2\left(\frac{\pi}{16}\right)\]\[A=\frac{3\pi a^2}{8}.\]
\[A=\frac{3\pi a^2}{8}\]
32

Area of a Generalised Hypocycloid

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 10 • 4 Marks

State Green’s theorem in the plane and use it to find the area of the hypocycloid

\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1.\]

Solution

Green’s theorem and parametrisation
Shaded generalised astroid with x intercept a and y intercept b
The intercepts of the generalised hypocycloid are \((\pm a,0)\) and \((0,\pm b)\).

By Green’s theorem,

\[A=\iint_Rdx\,dy=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]
\[A=\iint_Rdx\,dy\]\[=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]

The Cartesian equation of the generalised hypocycloid is

\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1.\]
\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=1.\]

Use the parametric form

\[x=a\cos^3t,\qquad y=b\sin^3t,\qquad0\le t\le2\pi.\]
\[x=a\cos^3t\]\[y=b\sin^3t\]\[0\le t\le2\pi.\]

Indeed,

\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}=\cos^2t+\sin^2t=1.\]
\[\left(\frac{x}{a}\right)^{2/3}+\left(\frac{y}{b}\right)^{2/3}\]\[=\cos^2t+\sin^2t\]\[=1.\]

Differentiating the parametric equations,

\[dx=-3a\cos^2t\sin t\,dt\]\[dy=3b\sin^2t\cos t\,dt.\]

Substituting the parametric values of \(x\), \(y\), \(dx\) and \(dy\),

\[A=\frac12\int_0^{2\pi}\left[(a\cos^3t)(3b\sin^2t\cos t)-(b\sin^3t)(-3a\cos^2t\sin t)\right]dt.\]
\[A=\frac12\int_0^{2\pi}\Big[(a\cos^3t)(3b\sin^2t\cos t)\]\[{}-(b\sin^3t)(-3a\cos^2t\sin t)\Big]dt.\]

Therefore,

\[A=\frac{3ab}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t+\sin^4t\cos^2t\right)dt.\]
\[A=\frac{3ab}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t\right.\]\[\left.{}+\sin^4t\cos^2t\right)dt.\]

Taking \(\sin^2t\cos^2t\) as the common factor,

\[A=\frac{3ab}{2}\int_0^{2\pi}\sin^2t\cos^2t\left(\cos^2t+\sin^2t\right)dt.\]
\[A=\frac{3ab}{2}\int_0^{2\pi}\sin^2t\cos^2t\]\[\qquad\cdot\left(\cos^2t+\sin^2t\right)dt.\]

Since \(\cos^2t+\sin^2t=1\),

\[A=\frac{3ab}{2}\int_0^{2\pi}\sin^2t\cos^2t\,dt.\]

The curve is symmetric in all four quadrants. Hence,

\[A=4\left(\frac{3ab}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]
\[A=4\left(\frac{3ab}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]

Therefore,

\[A=6ab\int_0^{\pi/2}\sin^2t\cos^2t\,dt.\]

Now,

\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt=\frac14\int_0^{\pi/2}\sin^22t\,dt=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt=\frac{\pi}{16}.\]
\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt\]\[=\frac14\int_0^{\pi/2}\sin^22t\,dt\]\[=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt\]\[=\frac{\pi}{16}.\]

Hence,

\[A=6ab\left(\frac{\pi}{16}\right)\]\[A=\frac{3\pi ab}{8}.\]
\[A=\frac{3\pi ab}{8}\]
33

Area of an Astroid

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 10 • 4 Marks

Using Green’s theorem, find the area enclosed by the astroid

\[x^{2/3}+y^{2/3}=a^{2/3}.\]

Solution

Green’s theorem and parametrisation
Shaded region enclosed by an astroid
The astroid is symmetric about both coordinate axes.

By Green’s theorem,

\[A=\iint_Rdx\,dy=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]
\[A=\iint_Rdx\,dy\]\[=\frac{1}{2}\oint_C(x\,dy-y\,dx).\]

The Cartesian equation of the astroid is

\[x^{2/3}+y^{2/3}=a^{2/3}.\]

Use the parametric form

\[x=a\cos^3t,\qquad y=a\sin^3t,\qquad0\le t\le2\pi.\]
\[x=a\cos^3t\]\[y=a\sin^3t\]\[0\le t\le2\pi.\]

Indeed,

\[x^{2/3}+y^{2/3}=a^{2/3}\cos^2t+a^{2/3}\sin^2t=a^{2/3}.\]
\[x^{2/3}+y^{2/3}\]\[=a^{2/3}\cos^2t+a^{2/3}\sin^2t\]\[=a^{2/3}.\]

Differentiating the parametric equations,

\[dx=-3a\cos^2t\sin t\,dt\]\[dy=3a\sin^2t\cos t\,dt.\]

Substituting the parametric values of \(x\), \(y\), \(dx\) and \(dy\),

\[A=\frac12\int_0^{2\pi}\left[(a\cos^3t)(3a\sin^2t\cos t)-(a\sin^3t)(-3a\cos^2t\sin t)\right]dt.\]
\[A=\frac12\int_0^{2\pi}\Big[(a\cos^3t)(3a\sin^2t\cos t)\]\[{}-(a\sin^3t)(-3a\cos^2t\sin t)\Big]dt.\]

Therefore,

\[A=\frac{3a^2}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t+\sin^4t\cos^2t\right)dt.\]
\[A=\frac{3a^2}{2}\int_0^{2\pi}\left(\cos^4t\sin^2t\right.\]\[\left.{}+\sin^4t\cos^2t\right)dt.\]

Taking \(\sin^2t\cos^2t\) as the common factor,

\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\left(\cos^2t+\sin^2t\right)dt.\]
\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\]\[\qquad\cdot\left(\cos^2t+\sin^2t\right)dt.\]

Since \(\cos^2t+\sin^2t=1\),

\[A=\frac{3a^2}{2}\int_0^{2\pi}\sin^2t\cos^2t\,dt.\]

The astroid is symmetric in all four quadrants. Hence,

\[A=4\left(\frac{3a^2}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]
\[A=4\left(\frac{3a^2}{2}\int_0^{\pi/2}\sin^2t\cos^2t\,dt\right).\]

Therefore,

\[A=6a^2\int_0^{\pi/2}\sin^2t\cos^2t\,dt.\]

Now,

\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt=\frac14\int_0^{\pi/2}\sin^22t\,dt=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt=\frac{\pi}{16}.\]
\[\int_0^{\pi/2}\sin^2t\cos^2t\,dt\]\[=\frac14\int_0^{\pi/2}\sin^22t\,dt\]\[=\frac18\int_0^{\pi/2}(1-\cos4t)\,dt\]\[=\frac{\pi}{16}.\]

Hence,

\[A=6a^2\left(\frac{\pi}{16}\right)\]\[A=\frac{3\pi a^2}{8}.\]
\[A=\frac{3\pi a^2}{8}\]
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