Integrate $f=\frac{Y}{X}\sqrt{4Z^2 + 1}$ over the portion of the paraboloid $Z=X^2+Y^2$ that lies above the rectangle with the following limits: $ 1\lt X\lt e , 0\lt y\lt 2 $ in the $X-Y$ plane."
I know that you use the formula, but i get stuck beyond that. Can someone show me a fully worked solution please on how to do this.
Many thanks
Please excuse the poor formatting. I don't know how to use this website properly just yet!
We are asked to integrate the function $$ f(x,y,z) = \frac{y}{x} \sqrt{4z^2 + 1} $$ over the surface $S$ defined by $$ S = \left\{(x,y,z)\mid 1 \leq x \leq e,\ 0 \leq y \leq 2,\ z = x^2 + y^2\right\} $$
Stop to make sure you understand: the surface $S$ is two-dimensional. If we are to integrate a function over $S$, it should break down to two variables. Since, on $S$, $z=x^2+y^2$, we substitute it into the function $f(x,y,z)$. So part of the integrand is $$f(x,y,z^2 + y^2) = \frac{y}{x}\sqrt{4x^2 + 4y^2+1}$$
Now remember the surface is not flat, but curved. That curvature is accounted for in the integral by integrating against $dS$ instead of $dA$. But since our surface is part of a graph, this is easy: $$ dS = \sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dx\,dy = \sqrt{(2x)^2 + (2y)^2 + 1}\,dx\,dy $$ Putting this together, we have $$ \begin{split} \iint_S f\,dS &= \int_1^e \int_0^2\frac{y}{x}\sqrt{4x^2 + 4y^2+1}\cdot \sqrt{(2x)^2 + (2y)^2 + 1}\,dx\,dy \\ &= \int_1^e \int_0^2\frac{y}{x}\left(4x^2 + 4y^2+1\right)\,dy\,dx \end{split} $$
Hopefully you can proceed from here. If not, reread your textbook's section on double and iterated integrals.