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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.

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  • $\begingroup$ Just wondering, how did you get the limits on integration? $\endgroup$ – JLL Apr 30 '15 at 2:19
  • $\begingroup$ As you have specified the surface you are integrating over, your integral does not converge. Do you also want a condition like $x^2+y^2\leq 1$? $\endgroup$ – Josh Burby Apr 30 '15 at 2:40
  • $\begingroup$ @JLL Sorry for the delay in response. I came up with the limits on integration with the idea that the vector projection of the parabolic dome onto the xy plane meant that $z = 0$. Thus, my idea was $z = 1 - x^2 - y^2$ meant that, for $z = 0, x$ and $y$ had to sum to something equal to 0. I was obviously incorrect. $\endgroup$ – Andrew Falanga May 1 '15 at 14:42
  • $\begingroup$ @JoshBurby Indeed, one of the things I've struggled with on this is understanding how these two elements work together for this. For example, in this problem, I now understand that $H(x,y,z)$ doesn't specify limits for integration, but is the function I'm integrating over the surface $z = 1 - x^2 - y^2$. Yes I know, an epiphany but for me it truly was. I kept trying to make the vector projection a part of $H(x,y,z)$ but it wasn't. It's simply the projection onto a 2 dimensional surface which enables me to do a double integral. $\endgroup$ – Andrew Falanga May 1 '15 at 14:46

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