Find the common area enclosed by the parabolas $4y^2 = 9x$; $3x^2 = 16y$.

> [!TLDR] > Finding the two intersection points of $4y^2=9x$ and $3x^2=16y$ and integrating the difference of the two curves over $x\in[0,4]$ gives a common area of $4$ square units. > > 1. Curve 1: $4y^2=9x$, i.e. $y=\dfrac32\sqrt x$ (taking the upper branch, since it bounds a region with the other parabola), a right-opening parabola with vertex at the origin. > 2. Curve 2: $3x^2=16y$, i.e. $y=\dfrac{3}{16}x^2$, an upward-opening parabola with vertex at the origin. > 3. **Points of intersection:** from curve 1, $x=\dfrac{4}{9}y^2$. Substituting into curve 2's form $x^2=\dfrac{16}{3}y$: > $\left(\dfrac49y^2\right)^2 = \dfrac{16}{3}y \Rightarrow \dfrac{16}{81}y^4 = \dfrac{16}{3}y \Rightarrow y^4 = 27y \Rightarrow y(y^3-27)=0$. > So $y=0$ or $y=3$. > 4. For $y=0$: $x=0$. For $y=3$: $x=\dfrac49(9)=4$. So the two parabolas meet at $(0,0)$ and $(4,3)$, enclosing a lens-shaped common region between them. > 5. For $0 6. Common (enclosed) area $\displaystyle A=\int_0^4\left[\dfrac32\sqrt x - \dfrac{3}{16}x^2\right]dx$. …

Find the common area enclosed by the parabolas 4y² = 9x; 3x² = 16y.

Question 69 of 96

4y^2 = 9x

Puducherry TnboardTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★

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Concept understanding — Area of a Bounded Plane Region by Integration

Building on the geometric meaning of $\int_a^b f(x)\,dx$ (Remarks under the limit-of-a-sum definition), this topic packages the vertical/horizontal-strip argument into ready-to-use area formulas.

Bounded by a curve, the $x$ -axis, and $x=a,\,x=b$ (vertical strips, height $=|y|$ , width $\Delta x$ ):

If $y=f(x)\ge0$ throughout $[a,b]$ (curve above the $x$ -axis): $A=\displaystyle\int_a^b y\,dx$ .
If $y=f(x)\le0$ throughout $[a,b]$ (curve below the $x$ -axis): $A=\displaystyle-\int_a^b y\,dx=\left|\int_a^b y\,dx\right|$ .
If $f$ changes sign on $[a,b]$ : split at the zeros $c_1,c_2,\ldots$ into subintervals of constant sign, apply the two rules above on each piece, and add the absolute values — $A=\left|\int_a^{c_1}f\right|+\left|\int_{c_1}^{c_2}f\right|+\cdots$ (never just $\int_a^b f\,dx$ , which can cancel positive against negative area).

Bounded by a curve, the $y$ -axis, and $y=c,\,y=d$ (horizontal strips, mirror image): $A=\displaystyle\int_c^d x\,dy$ if the curve lies to the right of the $y$ -axis ( $x\ge0$ ); $A=-\displaystyle\int_c^d x\,dy$ if to the left; split-and-add-absolute-values if it crosses.

Bounded between two curves. If $f(x)\ge g(x)$ on $[a,b]$ (an "upper" curve $U$ and "lower" curve $L$ ), the region between them and the ordinates $x=a,x=b$ has area

A=\int_a^b[f(x)-g(x)]\,dx=\int_a^b(y_U-y_L)\,dx.

The $y$ -axis mirror (curves $x=f(y)\ge x=g(y)$ , a "right" curve $R$ and "left" curve $L$ ) gives $A=\int_c^d(x_R-x_L)\,dy$ . …

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Finding the two intersection points of $4y^2=9x$ and $3x^2=16y$ and integrating the difference of the two curves over $x\in[0,4]$ gives a common area of $4$ square units.

Curve 1: $4y^2=9x$ , i.e. $y=\dfrac32\sqrt x$ (taking the upper branch, since it bounds a region with the other parabola), a right-opening parabola with vertex at the origin.
Curve 2: $3x^2=16y$ , i.e. $y=\dfrac{3}{16}x^2$ , an upward-opening parabola with vertex at the origin.
Points of intersection: from curve 1, $x=\dfrac{4}{9}y^2$ . Substituting into curve 2's form $x^2=\dfrac{16}{3}y$ : $\left(\dfrac49y^2\right)^2 = \dfrac{16}{3}y \Rightarrow \dfrac{16}{81}y^4 = \dfrac{16}{3}y \Rightarrow y^4 = 27y \Rightarrow y(y^3-27)=0$ . So $y=0$ or $y=3$ .
For $y=0$ : $x=0$ . For $y=3$ : $x=\dfrac49(9)=4$ . So the two parabolas meet at $(0,0)$ and $(4,3)$ , enclosing a lens-shaped common region between them.
For $0<x<4$ , compare the two branches: at $x=1$ , curve 1 gives $y=\tfrac32(1)=1.5$ and curve 2 gives $y=\tfrac{3}{16}(1)\approx0.19$ . So curve 1 ( $4y^2=9x$ ) lies above curve 2 ( $3x^2=16y$ ) throughout $(0,4)$ .
Common (enclosed) area $\displaystyle A=\int_0^4\left[\dfrac32\sqrt x - \dfrac{3}{16}x^2\right]dx$ . …

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Q.Find the common area enclosed by the parabolas 4y2=9x4y^2 = 9x4y2=9x; 3x2=16y3x^2 = 16y3x2=16y.

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Q.Find the common area enclosed by the parabolas $4y^2 = 9x$ ; $3x^2 = 16y$ .

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