Question about calculating the area underneath a "3d" curve.

Question

I'm trying to calculate the area underneath a curve after a $z$-component has been added.

Suppose we have the equation:

$$y = -x^4 - x^3 + 3x^2 -x + 4$$

on the interval $[-2.38, 1.76]$ (the roots of the function).

A wall with a constant height of $5$ feet will be modeled after the function. Calculate the area of this wall.

I first converted the curve into parametric form:

\begin{align} x(t) &= t \\ y(t) &= -t^4 - t^3 + 3t^2 -t + 4 \\ z(t) &= 5 \end{align}

I'm confused as to how the integral is set up. Is $z(t)$ the same as $f(x, y)$? And I'm also not sure if my $x$ values will be the bounds or if they need to be converted somehow.

Thanks

Answer 1

Hint: $$ A = h \, s = h \int\limits_a^b \sqrt{1+ (y'(x))^2}\,dx $$