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The problem:

$ \text{Set-up an integral for the surface area of the portion of the ellipsoid } 4x^2 + 9y^2 + z^2 = 64 \text{ that lies above the plane } z=-1 \text{. Do not simplify or evaluate the integral.}$

Professor's Answer:

$$ SA = \int_{-4}^4 \int_{ \sqrt{\frac{64}{9}-\frac{4x^2}{9}}}^{\sqrt{\frac{64}{9}-\frac{4x^2}{9}}} \sqrt{1+ (\frac{-4x}{\sqrt{64-4x^2-9y^2}})^2+ (\frac{-9y}{\sqrt{64-4x^2-9y^2}})^2} dA $$

$$ -\int_{-\sqrt{\frac{63}{4}}}^{\sqrt{\frac{63}{4}}} \int_{ \sqrt{\frac{63}{9}-\frac{4x^2}{9}}}^{\sqrt{\frac{63}{9}-\frac{4x^2}{9}}} \sqrt{1+ (\frac{-4x}{\sqrt{64-4x^2-9y^2}})^2+ (\frac{-9y}{\sqrt{64-4x^2-9y^2}})^2} dA $$ Where first double integral represents the whole ellipsoid and the second double integral represents the part below the surface.

I'm not exactly sure how he got much of his answer. The radical looks akin to the Jacobian when you take the area of function bounded by a plane. The second integration in the first double integral looks like he was solving for y to me but I'm having a hard time fitting the pieces together.

enter image description here

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  • $\begingroup$ Are you sure there was not a "$2$" in front of the first surface integral? $\endgroup$ – JG123 Aug 17 '19 at 15:25
  • $\begingroup$ yes, I've attached an image of his work $\endgroup$ – financial_physician Aug 17 '19 at 16:07
  • $\begingroup$ but if you think it makes more sense with a 2 and could explain how you got where you're at, might help us make some progress $\endgroup$ – financial_physician Aug 17 '19 at 16:23
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Well here is my rationale:

The ellipsoid in question can be represented by z=$\pm$$(\sqrt {64-4x^2-9y^2}$).

The "top half" is given by z=$(\sqrt {64-4x^2-9y^2}$), with $4x^2+9y^2$$\le$$64$.

In other words, we have that: -$(\frac{\sqrt {64-4x^2}}{3})$ $\le$y$\le$$(\frac{\sqrt {64-4x^2}}{3})$

and $-4$$\le$x$\le$$4$

Hence, the surface area of this "top half" will be the first surface integral that your professor wrote.

Now consider the portion of the ellipsoid above the plane $z=1$ (which will have the same surface area as the portion below the plane $z=-1$).

We now are computing a surface integral with the same surface z=$(\sqrt {64-4x^2-9y^2}$) but we have that $4x^2+9y^2$$\le$$63$. (This can be obtained by plugging $z=1$ into the expression for the ellipsoid in question).

In other words, we have that: -$(\frac{\sqrt {63-4x^2}}{3})$ $\le$y$\le$$(\frac{\sqrt {63-4x^2}}{3})$

and -$\frac{63}{4}$ $\le$ x $\le$ $\frac{63}{4}$.

Hence the surface area of the portion of the ellipsoid below the plane $z=-1$ will be the second surface integral that your professor wrote.

We know that the desired surface area will be: $SA_{ellipsoid}$-$SA_{S}$, where $S$ is the portion of the ellipsoid below the plane $z=-1$.

Does it not then make sense that the desired surface area will be 2(1st surface integral)-(2nd surface integral) as the 1st surface integral only gives the surface area of the top half of the ellipsoid?

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