Long Question 4

Solution:

$\begin{array}{l}x=y^2-1\to \left(1\right)\\ 3y=2x\\ x=\frac{3}{2}y\to \left(2\right)\\ \text{Substitute (2) into (1),}\\ \frac{3}{2}y=y^2-1\\ 2y^2-3y-2=0\\ \left(2y+1\right)\left(y-2\right)=0\\ y=-\frac{1}{2}\text{ or }y=2\end{array}$


$\begin{array}{l}\text{When }y=2,x=\frac{3}{2}\left(2\right)=3,Q=\left(3,2\right)\\ \\ I_1\left(\text{Volume of cone}\right)\\ =\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(3\right)}^2\left(2\right)\\ =6\pi {\text{ unit}}^3\\ \\ I_2\left(\text{Volume of the curve}\right)\\ \text{= }\pi {\displaystyle {\int }_1^2{x}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_1^2{\left(y^2-1\right)}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_1^2\left(y^4-2y^2+1\right)dy}\\ =\pi {\left[\frac{y^5}{5}-\frac{2y^3}{3}+y\right]}_1^2\\ =\pi \left[\left(\frac{2^5}{5}-\frac{2{\left(2\right)}^3}{3}+2\right)-\left(\frac{1^5}{5}-\frac{2{\left(1\right)}^3}{3}+1\right)\right]\\ =\pi \left(\frac{46}{15}-\frac{8}{15}\right)\\ =\frac{38}{15}\pi {\text{ unit}}^3\\ \\ \therefore \text{ Volume generated }=I_1-I_2\\ =6\pi -\frac{38}{15}\pi \\ =\frac{52}{15}\pi {\text{ unit}}^3\end{array}$