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**HallsofIvy** Here's a general concept: if you write the surface as $\displaystyle \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}$, then the two derivatives, $\displaystyle \vec{r}_u= x_u \vec{i}+ y_u\vec{j}+ z_u\vec{k}$ and $\displaystyle \vec{r}_v= x_v\vec{i}+ y_v\vec{j}+ z_v\vec{k}$ are tangent to the surface and their cross product (called the "fundamental vector product" of the surface), $\displaystyle \vec{r}_u\times\vec{r_v}$, is perpendicular to the surface and its length gives the "differential of surface area": $\displaystyle ||\vec{r}_u\times\vec{r}_v||dudv$.

Yes, you can use $\displaystyle y= cos(\theta)$, $\displaystyle z= sin(\theta)$, x= x with $\displaystyle \theta$ and x as parameters u and v. That is, $\displaystyle \vec{r}= x\vec{i}+ cos(\theta)\vec{j}+ sin(\theta)\vec{k}$ so $\displaystyle \vec{r}_x= \vec{i}$ and $\displaystyle \vec{r}_\theta= -sin(\theta)\vec{j}+ cos(\theta)\vec{k}$ so that the "fundamental vector product" is

$\displaystyle \vec{r}_x\times\vec{r}_\theta= -cos(\theta)\vec{i}- sin(\theta)\vec{k}$.

Its length is, of course, 1 so $\displaystyle dV= dxd\theta$. The function to be integrated is $\displaystyle z+ x^2y= sin(\theta)+ x^2cos(\theta)$. In the first octant, both y and z are positive so $\displaystyle \theta$ goes from 0 to $\displaystyle \pi/2$ and, of course, x goes from 0 to 1. The integral is

$\displaystyle \int_{x=0}^1\int_{\theta= 0}^{\pi/2} sin(\theta)+ x^2 cos(\theta) dxd\theta$.

By the way, if you click on a mathematical expression in any post, you can see the LaTex code used.