Answer:
[tex]Area=7\,\pi\,\sqrt{\frac{53}{4}}\,\,\frac{25}{2}[/tex]
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:
[tex]Area=\int\limits^b_a {2\,\pi\,y\,\sqrt{1+(\frac{dy}{dx} )^2} } \, dx[/tex]
and which in our case, we can obtain the following:
[tex]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} }[/tex]
Recall as well that [tex]0\leq x\leq 5[/tex], which gives us the limits of integration:
[tex]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}[/tex]
If we compare it with the geometry formula:
Lateral surface of cone = [tex]\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} }[/tex]
which is exactly the expression we calculated with the integral.
