Question

Find the lateral​ (side) surface area of the cone generated by revolving the line segment y equals seven halves x ​, 0 less than or equals x less than or equals 5​, about the​ x-axis. Check your answer with the following geometry formula. Lateral surface areaequalsone half times base circumference times slant height

Answer 1

`Area=7\,\pi\,\sqrt{(53)/(4)}\,\,(25)/(2)`

Explanation:

Let's use the integral formula for the surface area of revolution of the function y(x) around the x-axis, which is:

`Area=\int\limits^b_a {2\,\pi\,y\,\sqrt{1+((dy)/(dx) )^2} } \, dx`

and which in our case, we can obtain the following:

`y=(7)/(2) \,x\\(dy)/(dx) =(7)/(2) \\((dy)/(dx))^2=(49)/(4) \\\sqrt{1+((dy)/(dx))^2} =\sqrt{1+(49)/(4) } =\sqrt{(53)/(4) }`

Recall as well that
`0\leq x\leq 5`, which gives us the limits of integration:

`Area=\int\limits^b_a {2\,\pi\,y\,\sqrt{1+((dy)/(dx) )^2} } \, dx\\Area=\int\limits^5_0 {2\,\pi\,((7)/(2)\,x) \,\sqrt{(53)/(4) } } \, dx\\Area=7\,\pi\,\sqrt{(53)/(4)}\,\, \int\limits^5_0 {x} \, dx \\Area=7\,\pi\,\sqrt{(53)/(4)}\,\,(x^2)/(2) |\limits^5_0\\Area=7\,\pi\,\sqrt{(53)/(4)}\,\,(25)/(2)`

If we compare it with the geometry formula:

Lateral surface of cone =
`(1)/(2) \,\,(Base_(circ))\,\,(slant\,height)= (1)/(2) (2\,\pi\,(7)/(2) 5)\.(\sqrt{5^2+((35)/(2))^2 } =(7)/(2) \,\pi\,25\.\,\sqrt{(53)/(4) }`

which is exactly the expression we calculated with the integral.