Not getting surface integrals I have this problem from homework:
Integrate the given problem over the given surface.  $H(x,y,z)=x^2 \sqrt{5-4z}$ over the parabolic dome $z = 1-x^2-y^2, x \ge 0$
I used this formula from my book for surfaces $S$ given explicitly as the graph of $z=f(x,y)$.  $\int \int_S G(x,y,z)d\sigma = \int \int_R G(x,y,f(x,y)) \sqrt{f_x^2 + f_y^2 + 1}dxdy$.
So, what I have is something like this.  
$$
\begin{array}{rcl}
H(x,y,z) & = & x^2 \sqrt{5-4z} \\
         & = & x^2 \sqrt{1 + 4x^2 + 4y^2} \\
         & & \\
f(x,y) & = & 1 - x^2 - y^2 \\
   f_x & = & -2x \\
   f_y & = & -2y \\
   & & \\
\int \int_R H(x,y,z)\sqrt{f_x^2 + f_y^2 +1}dxdy & = & \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} x^2 (\sqrt{1 + 4x^2 + 4y^2})(\sqrt{1 + 4x^2 + 4y^2})dxdy \\
   & = & \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} x^2 + 4x^4 +4x^2y^2dxdy \\
   & = & \frac{43}{45}
\end{array}
$$
However, this is incorrect.  I'm not sure what I'm not getting.  I really need to have some insight.  The book shows this for the answer.
$$
\begin{array}{rcl}
\int \int_S x^2 \sqrt{5-4z}d\sigma & = & \int_{0}^{1} \int_{0}{2\pi} u^2cos(v)^2 \cdot \sqrt{4u^2 + 1} \cdot u\sqrt{4u^2+1}dvdu \\
      & = & \int_{0}^{1} \int_{0}^{2\pi} u^3(4u^2 + 1)cos(v)^2 dvdu \\
   & = & \frac{11\pi}{12}
\end{array}
$$
Please help me to see what it is I'm missing on setting up these things.  Thanks.
 A: I know this question was asked 5 years ago, but maybe I can be of help to future calc 3 students. I've spent 3 hours on this problem and finally figured it out...hopefully this helps!
Note: depending on your teacher's style, $u$ can be replaced with $r$ and $v$ can be replaced with $\theta$.
$$H(x,y,z)=x^2 \sqrt{5-4z},\quad z=1-x^2-y^2,\quad x≥0$$
$$\iint _S H(x,y,z)dσ= \iint_R H(u,v)|r_u×r_v |  dudv$$
Convert to cylindrical:
$$x=u\cos⁡ v,\quad y=u\sin⁡ v,\quad x^2+y^2=u^2,\quad 0≤v≤2\pi$$
$$z=1-(x^2+y^2 )=1-u^2,\quad 0≤z≤1,\quad 0≤u≤1$$
$$H(u,v)=u^2\cos^2⁡ v \sqrt{5-4(1-u^2 )}=u^2\cos^2 ⁡v\sqrt{4u^2+1}$$
Find determinant:
$$\vec{r}=u \cos ⁡v \hat{i} +u\sin⁡ v \hat{j}+(1-u^2 ) \hat{k}$$
$$\vec{r}_u=\cos⁡ v \hat{i}+ \sin⁡ v\hat{j}+(-2u)\hat{k}$$
$$\vec{r}_v=-u\sin ⁡v\hat{i}+u\cos ⁡v\hat{j}+0 \hat{k}$$
$$\vec{r}_u\times \vec{r}_v=(0-(-2u^2\cos ⁡v ))\hat{i}+(0-(-2u^2\sin⁡ v ))\hat{j}+(u \cos^2⁡v+u \sin ^2 ⁡v )\hat{k}$$
$$=2u^2\cos ⁡v \hat{i}+2u^2\sin ⁡v \hat{j}+u\hat{k}$$
$$|\vec{r}_u\times \vec{r}_v|=\sqrt{4u^4\cos^2 ⁡v+4u^4\sin^2⁡ v+u^2 }$$
$$=u\sqrt{4u^2+1}$$
Put it together:
$$H(u,v)|\vec{r}_u\times \vec{r}_v|=u^2  \cos^2⁡ v \sqrt{4u^2+1} u\sqrt{4u^2+1}$$
$$=\cos^2 ⁡v (4u^5+u^3 )$$
$$\iint _RH(u,v)|\vec{r}_u\times \vec{r}_v| dudv$$
$$=\int_0^{2\pi}\int _0^1\cos^2⁡v (4u^5+u^3 )  dudv$$
$$=\int_0^{2\pi}\cos^2⁡v \left({4\over6} u^6+{{1\over4} u}^4 \right)|^{1}_{0} dv$$
$$=\int_0^{2\pi}{11\over12}  \cos^2⁡v dv$$
$$=\int_0^{2\pi}\left({11\over12}\right) \left({1\over2}\right)(1+\cos⁡ 2v )dv$$
$$=\left.\left({11\over24} v+{1\over2}  \sin⁡2v \right)\right|^{2\pi}_0$$
$$={11\pi\over12}$$
