The area region bounded by $y=3x$, $x=5$ and $x$-axis

Question

Find the area region bounded by $y=3x$, $x=5$ and $x$-axis, hence find the volume revolved at $y$-axis $(360^\circ)$ and $y$-axis ($180^\circ$). How to find the volume and what it is mean by 'hence'. For the area i get the answer $75/2$ units. Please help since i cannot configure it other way around

Answer 1

Notice, plot the lines, $y=3x$, $x=5$ then one should find that the region bounded by the lines $y=3x$, $x=5$ & x-axis is a right triangle having vertices at $(0, 0)$, $(5, 0)$ & $(5, 15)$ hence, its area is given as $$A=\int_{x=0}^{x=5} y\ dx$$ $$=\int_{x=0}^{x=5} (3x)\ dx=3\left[\frac{x^2}{2}\right]_0^5=\color{red}{\frac{75}{2}}$$ Now, the volume of the solid obtained by rotating the curve (line) $y=3x$ about $y$-axis $$=\text{volume of complementary cylinder}-\text{volume of cone of rotation}$$ $$=\pi r^2l-\int_{y=0}^{y=15}\pi x^2\ dy $$ $$=\pi(5)^2(15)-\int_{y=0}^{y=15}\pi \left(\frac y3\right)^2\ dy $$

$$=375\pi-\frac{\pi}{9}\left[\frac{y^3}{3}\right]_{y=0}^{y=15} $$ $$=375\pi-125\pi=\color{red}{250\pi}$$