Question

Find the area of the region bounded by x² = 16y, y = 1, y = 4 and the y-axis in the first quadrant.

Answer 1

$$\begin{gathered} x^2=16y\mathrm{is}\mathrm{a}\mathrm{parabola},\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{O}\left(0,0\right)\mathrm{and}\mathrm{symmetrical}\mathrm{about}+\mathrm{ve}\mathit{}y-\mathrm{axis} \\ y=1\mathrm{is}\mathrm{line}\mathrm{parallel}\mathrm{to}x-\mathrm{axis}\mathrm{cutting}\mathrm{the}\mathrm{parabola}\mathrm{at}\left(-4,1\right)\mathrm{and}\left(4,1\right) \\ y=4\mathrm{is}\mathrm{line}\mathrm{parallel}\mathrm{to}x\mathit{}\mathrm{axis}\mathrm{cutting}\mathrm{the}\mathrm{parabola}\mathrm{at}\left(-8,1\right)\mathrm{and}\left(8,1\right) \\ \mathrm{Consider}\mathrm{a}\mathrm{horizontal}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|x\right|\mathrm{and}\mathrm{width}=dy \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|x\right|dy \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}y=1\mathrm{to}y=4 \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}\mathrm{enclosed}\mathrm{by}y=1\mathrm{and}y=4\mathrm{is}\mathrm{the}\mathrm{shaded}\mathrm{area} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_1^4\left|x\right|dy \\ \Rightarrow A={\int }_1^{4}xdy\left[As,x>0,\left|x\right|=x\right] \\ \Rightarrow A={\int }_1^4\sqrt{16y}dy \\ \Rightarrow A=4{\int }_1^4\sqrt{y}dy \\ \Rightarrow A=4{\left[\frac{y^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_1^4 \\ \Rightarrow A=\frac{8}{3}\left[4^{\frac{3}{2}}-1^{\frac{3}{2}}\right] \\ \Rightarrow A=\frac{8}{3}\times 7=\frac{56}{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{enclosed}\mathrm{by}\mathrm{parabola}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}\mathrm{and}y=1\mathrm{and}y=4\mathrm{is}\frac{56}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$