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Exercise 9.8 · Q10

Q.Find the area of the region common to the circle x2+y2=16x^2+y^2=16 and the parabola y2=6xy^2=6x.

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Find the single valid intersection, determine which curve binds the width on each xx-range by testing a point, integrate each piece, then double using the xx-axis symmetry.

Step 1. Find the intersection. Substituting y2=6xy^2=6x into x2+y2=16x^2+y^2=16: x2+6x−16=0⇒(x+8)(x−2)=0⇒x=2x^2+6x-16=0\Rightarrow(x+8)(x-2)=0\Rightarrow x=2 (the root x=−8x=-8 is rejected since y2=6xy^2=6x needs x≥0x\ge0). At x=2x=2, y2=12⇒y=±23y^2=12\Rightarrow y=\pm2\sqrt3.

Step 2. Identify which curve bounds the common region on each piece. For x∈[0,2]x\in[0,2]: at x=1x=1, parabola gives ∣y∣=6≈2.45|y|=\sqrt6\approx2.45, circle gives ∣y∣=15≈3.87|y|=\sqrt{15}\approx3.87 — the parabola is the tighter (binding) boundary. For x∈[2,4]x\in[2,4]: at x=3x=3, parabola gives ∣y∣=18≈4.24|y|=\sqrt{18}\approx4.24, circle gives ∣y∣=7≈2.65|y|=\sqrt7\approx2.65 — the circle is now the binding boundary. So, using symmetry about the xx-axis,

A=2[∫026x dx+∫2416−x2 dx].A=2\left[\int_0^{2}\sqrt{6x}\,dx+\int_{2}^{4}\sqrt{16-x^2}\,dx\right].

Step 3. First integral (area under the parabola).

∫026x dx=6[23x3/2]02=6⋅23⋅22=4123=833.\int_0^2\sqrt{6x}\,dx=\sqrt6\left[\dfrac23x^{3/2}\right]_0^2=\sqrt6\cdot\dfrac23\cdot2\sqrt2=\dfrac{4\sqrt{12}}3=\dfrac{8\sqrt3}3.

Step 4. Second integral (area under the circle), using ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa\displaystyle\int\sqrt{a^2-x^2}\,dx=\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}2\sin^{-1}\dfrac xa with a=4a=4.

∫2416−x2 dx=[x216−x2+8sin⁡−1x4]24.\int_2^4\sqrt{16-x^2}\,dx=\left[\dfrac x2\sqrt{16-x^2}+8\sin^{-1}\dfrac x4\right]_2^4. …

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