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

Answer:

$Area=7\,\pi\,\sqrt{\frac{53}{4}}\,\,\frac{25}{2}$

Step-by-step explanation:

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

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

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

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

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

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

If we compare it with the geometry formula:

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

which is exactly the expression we calculated with the integral.