Volume by integration

A.) To get the volume and centroid of the solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the x-axis, the solution look like this: We need a figure. It looks like this: Adding some finer details, it looks like this: The chord that passes through the parabola and perpendicular to the axis is the latus rectum of a parabola and its length is $4a$ if the equation of parabola is $y^2 = 4ax$ as shown below. The volume of solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the x-axis can be get by the disk method. One disk element has the volume $dV$ and its volume is $\pi r^2 dh$, where $dV$ is the volume element, $r$ is the radius and $dh$ is the height element. In this case: Now getting the volume of the figure above: For one volume element for the figure above, its volume is: $$dV = \pi r^2 dh$$ If we are going to add many volume elements to create a solid figure, the volume becomes: $$V = \int \pi r^2 dh$$ In this case, the volume of the solid generated above is: $$V = \int \pi y^2 dx$$ Now getting its volume: $$V = \int \pi y^2 dx$$ $$V = \int \pi (8x) dx$$ $$V = \int_0^2 \pi (8x) dx$$ $$\color{green}{V = 16\pi \space cubic \space units}$$ Now getting its centroid; the centroid of the figure above can be seen on the x-axis because of symmetry as shown below: Because of figure's symmetry, there is only one coordinate where the centroid of the figure lies. In this case, the centroid lies on the x-axis. The formula to get the centroid of the figure is: $$V\bar x = \int x_c dV$$ Then, in this case, the centroid of the figure above is: $$16\pi \bar x = \int x dV$$ Then: $$16\pi \bar x = \int x (\pi y^2 dx)$$ $$16\pi \bar x = \int_0^2 x (\pi (8x) dx)$$ $$16\pi \bar x = \int_0^2 8\pi x^2 dx $$ $$16\pi \bar x = \frac{64}{3} \pi$$ $$\color{green}{\bar x = \frac{4}{3}}$$ B.) To get the volume and centroid of the solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the y-axis, the solution look like this: We need a figure. It looks like this: Adding some finer details, it looks like this: The chord that passes through the parabola and perpendicular to the axis is the latus rectum of a parabola and its length is $4a$ if the equation of parabola is $y^2 = 4ax$ as shown below. The volume of solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the y-axis can be get by the ring method. To get the volume of the ring, subtract volume of one big ring by volume of one small ring as shown below: One ring element has the volume $dV$ and its volume is $\pi (r_2^2 - r_1^2) dh$, where $dV$ is the volume element, $r_2$ is the radius of big cylinder, $r_1^2$ is the radius of one small cylinder and $dh$ is the height element. In this particular problem: Now getting the volume of the figure above: For one volume element for the figure above, its volume is: $$dV = \pi (r_2^2 - r_1^2) dh$$ If we are going to add many volume elements to create a soid figure, the volume becomes: $$V = \int \pi (r_2^2 - r_1^2) dh$$ In this case, the volume of the solid generated above is: $$V = \int \pi ((2)^2 - (x)^2) dy$$ Now getting its volume: $$V = \int_0^4 \pi ((2)^2 - (x)^2) dy$$ $$V = \int_0^4 \pi (4 - x^2) dy$$ $$V = \int_0^4 \pi (4 - \left(\frac{y^4}{64}\right)) dy$$ $$\color{green}{V = \frac{64}{5}\pi \space cubic \space units}$$ Now getting its centroid; the centroid of the figure above can be seen on the y-axis because of symmetry as shown below: Because of figure's symmetry, there is only one coordinate where the centroid of the figure lies. In this case, the centroid lies on the y-axis. The formula to get the centroid of the figure is: $$V\bar y = \int y_c dV$$ Then, in this case, the centroid of the figure above is: $$\frac{64}{5}\pi \bar y = \int y dV$$ Then looking at the picture above, its obvious that the centroid of the figure below is $\color{green}{y_c = 0}$ C.) To get the volume and centroid of the solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the line $x = 2$ (the equation of the latus rectum of the curve $y^2 = 8x$), the solution look like this: We need a figure. It looks like this: The volume of solid of revolution generated by the curve $y^2 = 8x$, the x-axis and the latus rectum of $y^2 = 8x$ when its area rotates about the $x = 2$ can be get by the ring method. To get the volume of the ring, subtract volume of one big ring by volume of one small ring as shown below: One ring element has the volume $dV$ and its volume is $\pi (r_2^2 - r_1^2) dh$, where $dV$ is the volume element, $r_2$ is the radius of big cylinder, $r_1^2$ is the radius of one small cylinder and $dh$ is the height element. In this particular problem: Now getting the volume of the figure above: For one volume element for the figure above, its volume is: $$dV = \pi (r_2^2 - r_1^2) dh$$ If we are going to add many volume elements to create a soid figure, the volume becomes: $$V = \int \pi (r_2^2 - r_1^2) dh$$ In this case, the volume of the solid generated above is: $$V = \int \pi ((2)^2 - (x)^2) dy$$ Now getting its volume: $$V = \int_-4^4 \pi ((2)^2 - (x)^2) dy$$ $$V = \int_-4^4 \pi (4 - x^2) dy$$ $$V = \int_-4^4 \pi (4 - \left(\frac{y^4}{64}\right)) dy$$ $$\color{green}{V = \frac{128}{5}\pi \space cubic \space units}$$ Now getting its centroid; the centroid of the figure above can be seen on the x-axis because of symmetry as shown below: Because of figure's symmetry, there is only one coordinate where the centroid of the figure lies. In this case, the centroid lies on the x-axis. The formula to get the centroid of the figure is: $$V\bar y = \int y_c dV$$ Then, in this case, the centroid of the figure above is: $$\frac{128}{5}\pi \bar y = \int x dV$$ Then looking at the picture above, its obvious that the centroid of the figure below is $\color{green}{(x,y) = (2,0)}$ Alternate solutions are encouraged.....