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.
Best Answer
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: $$\vec{r}=u \cos v \hat{i} +u\sin v \hat{j}+(1-u^2 ) \hat{k}$$
$$\vec{r}_u=\cos v \hat{i}+ \sin v\hat{j}+(-2u)\hat{k}$$ $$\vec{r}_v=-u\sin v\hat{i}+u\cos v\hat{j}+0 \hat{k}$$
$$\vec{r}_u\times \vec{r}_v=(0-(-2u^2\cos v ))\hat{i}+(0-(-2u^2\sin v ))\hat{j}+(u \cos^2v+u \sin ^2 v )\hat{k}$$ $$=2u^2\cos v \hat{i}+2u^2\sin v \hat{j}+u\hat{k}$$
$$|\vec{r}_u\times \vec{r}_v|=\sqrt{4u^4\cos^2 v+4u^4\sin^2 v+u^2 }$$
$$=u\sqrt{4u^2+1}$$
Put it together: $$H(u,v)|\vec{r}_u\times \vec{r}_v|=u^2 \cos^2 v \sqrt{4u^2+1} u\sqrt{4u^2+1}$$ $$=\cos^2 v (4u^5+u^3 )$$
$$\iint _RH(u,v)|\vec{r}_u\times \vec{r}_v| dudv$$ $$=\int_0^{2\pi}\int _0^1\cos^2v (4u^5+u^3 ) dudv$$
$$=\int_0^{2\pi}\cos^2v \left({4\over6} u^6+{{1\over4} u}^4 \right)|^{1}_{0} dv$$ $$=\int_0^{2\pi}{11\over12} \cos^2v dv$$ $$=\int_0^{2\pi}\left({11\over12}\right) \left({1\over2}\right)(1+\cos 2v )dv$$ $$=\left.\left({11\over24} v+{1\over2} \sin2v \right)\right|^{2\pi}_0$$ $$={11\pi\over12}$$