Question

find the area of surface obtained by rotating the curves x=3t+5, y=4t+7 0<= t<=1 about x-axis

Answer 1

beging by maka a graph of the functions

Answer 2

i thik u meant the volume instead of area

Answer 3

no its the area of the polar coordinates

Answer 4

@terenzreignz

Answer 5

It's a good thing these parametric equations are both linear. Perhaps they can be joined into a single equation involving only x and y?

Answer 6

how can i do that?

Answer 7

Solve for t in both equations.

Answer 8

I got t=-5/3 and t=-7/4

Answer 9

Notice that we can find the rectangular equation by solving for 't' from the x = 3t + 5 and plugging that value in for 't' in y = 4t + 7:\[\bf x=3t+5 \implies t = \frac{x-5}{3} \implies y = 4\left( \frac{ x -5}{ 3 } \right)+7=\frac{4x+1}{3}\]Now since we are given that we must solve for \(
\bf 0 \le t \le 1\) about the x-axis, this is the same as saying that we must solve for \(
\bf 5 \le x \le 8\) rotated about the x-axis, since x = 3t + 5. Now that we have stated the function explicitly in terms of 'x', we see that when we rotate it about the x-axis, we get something like this:When you rotate this region around the x-axis, the cross-sections will be circles with a radius of:\[\bf Radius = f(x)=\frac{4x+1}{3}\]To find the surface area of this solid, we will be finding the circumference of each circle and multiplying it by a small band:You can think of the "infinitesimal hypotenuse" to be something similar to the slant height of a cone when one is calculating a cone's surface area. We are able to arrive at the formula:\[\bf S.A=\int\limits_{a}^{b}2\pi r \sqrt{1+\left( \frac{ dy }{ dx } \right)^2}dx=\int\limits_{5}^{8}2\pi \left( \frac{ 4x+1 }{ 3 } \right) \sqrt{1+\left( \frac{ 4 }{ 3 } \right)^2}dx\]Can you evaluate the integral?

@Brittny