Find the volume of the solid obtained by rotating the region bounded by
$$y=2x+6,\hspace{10mm} y=0$$
about the $y$-axis.
I have tried
$$\int\limits_0^6\pi\left(\frac{y-6}{2}\right)^2\,dy$$
but the answer is coming out incorrect.
[Math] Find the volume of the solid obtained by rotating about the y – axis
calculusvolume
Best Answer
I think your integral is almost correct (observe that when putting $\;x\;$ as function of $\;y\;$ gives you the volume of a function between the zero axis and the function!). What did you get?
$$\pi\int_0^6\left[0^2-\frac{(y-6)^2}4\right]dy=-\left.\frac\pi4\frac13(y-6)^3\right|_0^6=\frac\pi{12}216=18\pi$$
Another way to see it: you're rotating a straight angle triangle, with leg of length $\;3\;$, around its other leg of length $\;6\;$, so you can "place" this triangle on the $\;x\;$ axis at vertices $\;(0,0)\,,\,\,(6,0)\,,\,\,(6,3)\;$ , and thus the hypotenuse is given by the line $\;y=\frac12x\;$ , and doing the volume-by-rotation integral we get
$$\pi\int_0^6\frac{x^2}4dx=\frac\pi{12}6^3=18\pi$$