volume of solid generated by the regin bounded by curve $y=\sqrt{x},y=\frac{x-3}{2},y=0$ about $x$ axis

Question

Using sell method to find the volume of solid generated by revolving the region bounded by $$y=\sqrt{x},y=\frac{x-3}{2},y=0$$ about $x$ axis, is (using shell method)

What I try:

Solving two given curves $$\sqrt{x}=\frac{x-3}{2}\Longrightarrow x^2-10x+9=0$$

We have $x=1$ (Invalid) and $x=9$ (Valid).

Put $x=9$ in $y=\sqrt{x}$ we have $y=3$

Now Volume of solid form by rotation about $x$ axis is

$$=\int^{9}_{0}2\pi y\bigg(y^2-2y-3\bigg)dy$$

Is my Volume Integral is right? If not then how do I solve it? Help me please.

Answer 1

So I would instead split this up into two integrals:

$$\pi\int_0^3{(\sqrt{x})^2}dx + \pi\int_3^9{(\sqrt{x})^2-\left(\frac{x-3}{2}\right)^2}dx$$.

Using the shell method:

$$\int_0^3{2\pi y(2y+3-y^2)}dy$$.

Answer 2

Divide the region bounded by the curves $y_1=\sqrt x$ & $y_2=\frac{x-3}{2}$ from $x=0$ to $x=9$ into two parts. One part is bounded by the curve $y_1=\sqrt x$ from $x=0$ to $x=9$ & revolved around x-axis. Other part is bounded by the curve $y_2=\frac{x-3}{2}$ from $x=3$ to $x=9$ & revolved around x-axis

The required volume of bounded region revolved around x-axis is given as
$$\int_0^{9}\pi y_1^2dx-\int_{3}^{9} \pi y_2^2\ dx$$ $$=\int_0^{9}\pi (\sqrt x)^2dx-\int_{3}^{9} \pi \left(\frac{x-3}{2}\right)^2\ dx$$