Answer:
Step-by-step explanation:
Let take a look at the given function y = 4x - 1 whose point is located between (1,3) and (4,15) on the graph.
Here, the function of y is non-negative. Now, expressing y in terms of x in y = 4x- 1
4x = y + 1
[tex]x = \dfrac{y+1}{4}[/tex]
[tex]x = \dfrac{1}{4}y + \dfrac{1}{4}[/tex]
By integration, the required surface area in the revolve is:
[tex]S = \int^{15}_{ 3} 2 \pi g (y) \sqrt{1+g'(y^2) \ dy }[/tex]
where;
g(y) = [tex]x = \dfrac{1}{4}y + \dfrac{1}{4}[/tex]
∴
[tex]S = \int^{15}_{ 3} 2 \pi \Big( \dfrac{1}{4}y + \dfrac{1}{4}\Big) \sqrt{1+\Bigg(\Big( \dfrac{1}{4}y + \dfrac{1}{4}\Big)'\Bigg)^2 \ dy }[/tex]
[tex]S = \dfrac{1}{2} \pi \int^{15}_{ 3} (y+1) \sqrt{1+\Bigg(\Big( \dfrac{1}{4}\Big ) \Bigg)^2 \ dy } \\ \\ \\ S = \dfrac{1}{2} \pi \int^{15}_{ 3} (y+1) \dfrac{\sqrt{17}}{4} \ dy[/tex]
[tex]S = \dfrac{\sqrt{17}}{8} \pi \int^{15}_{ 3} (y+1) \ dy[/tex]
[tex]S = \dfrac{\sqrt{17} \pi}{8} (\dfrac{1}{2}(y+1)^2)\Big|^{15}_{3} \\ \\ S = \dfrac{\sqrt{17} \pi}{8} (\dfrac{1}{2}(15+1)^2-\dfrac{1}{2}(3+1)^2 ) \\ \\ S = \dfrac{\sqrt{17} \pi}{8} *120 \\ \\\mathbf{ S = 15 \sqrt{17}x}[/tex]
