Find the Area Between Circle and Parabola Using Integrals

Definite integral math problem

The figure shows two parabolas and a circle. The circle is passing through the vertex of the parabolas and the centre at the origin. If parabolas are y = x²/4 – 2 and y = –x²/4 + 2 then find the area between circle and parabola using integration

If we need to find the area using integration, then we need the intersection points of the curves

Area between circle and parabola

If we need to find the area using integration, then we need the intersection points of the curves

Area between circle and parabola

From figure

to find the intersecting points of the parabolas we can equate y = x²/4 – 2 and y = –x²/4 + 2 (we can also equate with x-axis)

x²/4 – 2 = –x²/4 + 2

⇒ x²/2 = 4

⇒ x² = 8

⇒ x = 2√2

then, y = 0

So intersecting points are (2√2, 0), (-2√2, 0)

that is A = (-2√2, 0) and B = (2√2, 0)

Point C is the intersecting of the parabola and circle (C also belong to y-axis)

y = –x²/4 + 2

⇒ y = 0 + 2

that is C = (0, 2) which means radius of the circle = 2 cm

so equation of the circle is x² + y² = 4

The above figure is symmetrical through the x-axis so we can find the area between the circle and parabola is twice the upper shaded area

e above figure is symmetrical through the x-axis so we can find the area between the circle and parabola is twice the upper shaded area

Now we can find the area of the parabola and the area of the semicircle separately and the difference between them is the upper shaded area

Area of the parabola

From figure area of the parabola is

\begin{aligned}
\int_{-2\sqrt2}^{2\sqrt2}y \ dx&=\int_{-2\sqrt2}^{2\sqrt2}\bigg(\frac{x^2}{4}-2\bigg) \ dx \\
\\
&=\bigg[\frac{x^3}{12}-2x\bigg]_{-2\sqrt2}^{2\sqrt2} \\
\\
&=\bigg[\frac{16\sqrt2}{12}-4\sqrt2\bigg]-\bigg[\frac{-16\sqrt2}{12}+4\sqrt2\bigg]\\
\\
&= \frac{4\sqrt2}{3}-4\sqrt2 + \frac{4\sqrt2}{3}-4\sqrt2\\
\\
&= \frac{8\sqrt2}{3}-8\sqrt2\\
\\
\int_{-2\sqrt2}^{2\sqrt2}y \ dx&= -\frac{16\sqrt2}{3}\\
\\
\end{aligned}

So we got the area of the parabola is (16√2)/3 unit² (Area is always positive)

Area of the semicircle

\begin{aligned}
\int_{-2}^{2}y \ dx&=\int_{-2}^{2}\sqrt{4-x^2} \ dx \\
\\
&=\bigg[\dfrac{4\sin^{-1}\left(\frac{x}{2}\right)+x\sqrt{4-x^2}}{2}\bigg]_{-2}^{2}\\
\\
&=\bigg[\dfrac{4\sin^{-1}1+2\sqrt{4-4}}{2}\bigg]-\bigg[\dfrac{4\sin^{-1}(-1)-2\sqrt{4-4}}{2}\bigg]\\
\\
&=\bigg[\dfrac{4\sin^{-1}1}{2}\bigg]-\bigg[\dfrac{4\sin^{-1}(-1)}{2}\bigg]\\
\\
&=\bigg[2 \times \frac{\pi}{2}\bigg]-\bigg[2 \times \frac{-\pi}{2}\bigg]\\
\\
\int_{-2}^{2}y \ dx&=2\pi
\end{aligned}

So the area of the semicircle is 2π unit²

The area of the upper shaded region is (16√2)/3 – 2π unit²

thus the area between the parabola and circle is (32√2)/3 – 4π unit²

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