Find the area of the region bounded by the curves |
\[{{y}^{2}}=4ax\] and \[{{x}^{2}}=4ay.\] |
OR |
Using integration, find the area of the triangular region whose sides have the equation \[y=2x+1,\]\[y=3x+1\] and x = 4. |
Answer:
The given curves are \[{{y}^{2}}=4ax\] [right parabola] ?(i) and \[{{x}^{2}}=4ay\] [upward parabola] ...(ii) On squaring both sides of Eq. (i), we get \[{{({{y}^{2}})}^{2}}={{(4ax)}^{2}}\] \[\Rightarrow \] \[{{y}^{4}}=16{{a}^{2}}{{x}^{2}}\] \[\Rightarrow \] \[{{y}^{4}}=16{{a}^{2}}(4ay)\] [from Eq. (ii)] \[\Rightarrow \] \[y({{y}^{3}}-64{{a}^{3}})=0\] \[\Rightarrow \] y = 0, \[{{y}^{3}}={{(4a)}^{3}}\] \[\Rightarrow \] y = 0, y = 4a When \[y=0,\] then x = 0 When y = 4a, then x = 4a Thus, the given parabolas intersect each other at 0(0, 0) and A(4a, 4a). Then, the shaded part in the figure is the required area. For the curve \[{{y}^{2}}=4ax,\] \[y=2\sqrt{a}\sqrt{x}=f(x)\,\,(\text{say)}\] and for the curve \[{{x}^{2}}=4ay,\] \[y=\frac{{{x}^{2}}}{4a}=g(x)(\text{say})\] \[\therefore \] Required area = Area of shaded region OACO \[=\int_{0}^{4a}{[f(x)-g(x)]}\,dx\] \[=\int_{0}^{4a}{2\sqrt{a}\sqrt{x}dx-\int_{0}^{4a}{\frac{{{x}^{2}}}{4a}dx}}\] \[=2\sqrt{a}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{4a}-\frac{1}{4a}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4a}\] \[=\frac{4}{3}\sqrt{a}{{(4a)}^{3/2}}-\frac{1}{12}{{(4a)}^{3}}\] \[=\frac{32{{a}^{2}}}{3}-\frac{16{{a}^{2}}}{3}=\frac{16{{a}^{2}}}{3}\] sq units OR We have to find the triangular region area whose equations of sides of triangle are given as \[Y=2x+1\] ?(i) \[Y=3x+1\] ?(ii) and x = 4 ?(iii) The line \[y=2x+1\]passes through the points (0, 1) and (1, 3) and the line \[y=3x+1\] passes through the points (0, 1) and (1, 4). Now, subtracting Eq. (i) from Eq. (ii), we get x = 0 \[\Rightarrow \] \[y=2(0)+1=1\] So, lines \[y=2x+1\] and \[y=3x+1\] meet at the point A (0, 1). Again, solving Eqs. (ii) and (iii), we get \[y=3x+1\] \[\Rightarrow \]\[y=3(4)+1\] \[[\because \,\,\,x=4]\] \[\Rightarrow \] \[y=12+1=13\] So, lines \[y=3x+1\] and x = 4 meet at the point C (4, 13). On solving Eqs. (i) and (iii), we get \[y=2x+1\] \[\Rightarrow \]\[y=2(4)+1\] \[[\because \,\,\,x=4]\] \[y=8+1=9\] So, lines \[y=2x+1\] and x = 4 meet at the point B (4, 9). Graph of the required region is given below: We have to find the area of \[\Delta \,ABC\] shaded above. Hence, required area of \[\Delta \,ABC\] \[=\text{Area}\,\,\text{of}\,\,AOHC-\text{Area}\,\,\text{of}\,\,AOHB\] \[=\int_{0}^{4}{(3x+1)\,dx-\int_{0}^{4}{(2x+1)\,dx}}\] \[=\left[ \frac{3{{x}^{2}}}{2}+x \right]_{0}^{4}-\left[ \frac{2{{x}^{2}}}{2}+x \right]_{0}^{4}\] \[=\left[ \frac{3(16)}{2}+4 \right]-[16+4]\] \[=24+4-16-4=8\,\text{sq}\,\,\text{units}\]
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