Evaluating $\iint_\Sigma \lvert xyz\rvert dS$

Question

How to get the value of $$\iint_\Sigma \lvert xyz\rvert dS$$, where $\Sigma$ is the finite surface cut from the elliptic paraboloid $z=x^2+y^2$ by the plane $z=1$?

According to the key, the result should be $\frac{125\sqrt{5}-1}{420}$. I cannot continue after I get $$∫_0^1 r^5 \sqrt{1+4r^2}dr$$

Thank you in advance.

Answer 1

(By the way, you are off by a factor of $2$.)

$$J:=2\int_0^1 r^5\sqrt{1+4r^2}\>dr=\int_0^1(r^2)^2\sqrt{1+4r^2}\>2rdr=\int_0^1u^2\sqrt{1+4u}\>du\ .$$ Now "substitute the square root", i.e., such that $\sqrt{1+4u}=v$. Then $$u={1\over4}(v^2-1),\quad du={1\over2}vdv\ ,$$ and we obtain $$J=\int_1^{\sqrt{5}}{1\over16}(v^2-1)^2v\>{1\over2}v\>dv=\ldots={125\sqrt{5}-1\over 420}\ .$$

Answer 2

If you differentiate $$r^4(1+4r^2)^{3/2}$$ you get $$12r^5\sqrt{1+4r^2}+4r^3(1+4r^2)^{3/2}=28r^5\sqrt{1+4r^3}+4r^3\sqrt{1+4r^2}$$. Similarly, differentiating $$r^2(1+4r^2)^{3/2}$$ gives $$12r^3\sqrt{1+4r^2}+2r(1+4r^2)^{3/2}=20r^3\sqrt{1+4r^2}+2r\sqrt{1+4r^2}$$ and differentiating $$(1+4r^2)^{3/2}$$ gives $$12r\sqrt{1+4r^2}$$.

So evidently we need $$\left(\frac{1}{28}r^4-\frac{1}{140}r^3+\frac{1}{840}\right)(1+4r^2)^{3/2}$$. Evaluating between $0$ and $1$ gives $\frac{125\sqrt5-1}{840}$.

For the original problem of finding the surface area there is probably also a factor $4\pi$ and maybe you need to add the top surface of the disk.

Answer 3

Here is a route by trigonometric substitution.

By the substitution: $$r=\frac12 \tan t,dr=\frac12 \sec^2 t dt$$

We have

\begin{align} \int r^5 \sqrt{1+4r^2}dr&=\int \frac1{32}\tan^5 t\cdot\sec t\cdot\frac12 \sec^2 t dt\\ &=\frac1{64}\int \tan^5 t \sec^3t dt\\ &=\frac1{64}\int\tan^4t\sec^2t d(\sec t)\\ &=\frac1{64}\int (\sec^2t-1)^2\sec^2t d(\sec t) \end{align}

Expand the term and integrate to get the desired result.