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One of my friend's students attempted to find the volume of a paraboloid using spherical coordinates. I'm trying to see if there's actually a way to finish it. I have been able to find the triple integral which describes the volume of the paraboloid in spherical coordinates and the part that I need help integrating is: $$\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx.$$

I know that it is equal to $4$ from Wolfram Alpha plus computing the same integral in cylindrical coordinates. Is there a way to finish this from here?

Someone asked for the original problem so here it is. Compute the volume of the region bounded by $z = 1 + x^2 + y^2$ and $z \leq 5$. This is the same as computing the region bounded by $z = 4 - x^2 - y^2$ and the $xy$-plane. Then from there I get the triple integral: $$\int_{0}^{2\pi}\int_0^{\pi/2}\int_0^{\frac{-\cos{\phi} + \sqrt{\cos^2{(\phi)} + 16\sin^2{(\phi)}}}{2\sin^2{(\phi)}}} \rho^2\sin{\phi}d\rho d\phi d\theta.$$

Of course the integral up top that I care about I just substituted the $\phi$ for an $x$ when I originally stated it by itself.

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    $\begingroup$ Would you mind stating the original problem so we can verify this integral here is correct? $\endgroup$
    – help
    Commented Nov 12, 2019 at 2:43
  • $\begingroup$ Okay I have updated the question. $\endgroup$
    – EgoKilla
    Commented Nov 12, 2019 at 3:07

2 Answers 2

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Let $t=\cos x$, we have \begin{align} I&=\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx \\ &= \frac{1}3\int_0^1 dt \left(\frac{-t+\sqrt{t^2+16(1-t^2)}}{2(1-t^2)}\right)^3 \end{align}

Let $z$ be the larger root of the equation $(1-t^2)z^2+tz-4=0$, we have $z=2$ at $t=0$ and $z=4$ at $t=1$. Let $y=tz$, we have $z^2=y^2-y+4\Rightarrow zdz=(y-1/2)dy$, $y=0$ at $t=0$ and $y=4$ at $t=1$, then

\begin{align} I&=\frac{1}3\int_0^1 dt z(t)^3 \\ &=\frac{1}3\left(z(t)^3t|_0^1-3\int_2^4t(z)z^2dz\right) \\ &=\frac{64}{3}-\int_0^4y\left(y-\frac{1}{2}\right)dy \\ &=\frac{64}{3}-\frac{64}{3}+\frac{16}{4}=4 \end{align}

Also a simple solution to your original problem:

area $A(z)=\pi r(z)^2=\pi(x^2+y^2)=\pi(z-1)$, \begin{align} V&=\int A(z) dz \\ &=\int_1^5\pi(z-1)dz \\ &=8\pi \end{align}

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  • $\begingroup$ How is $sin^2(x) = (1 + t^2)$? $\endgroup$
    – EgoKilla
    Commented Nov 12, 2019 at 3:11
  • $\begingroup$ Sorry, it's a typo. $\endgroup$ Commented Nov 12, 2019 at 3:12
  • $\begingroup$ How do you get from $\frac{1}{3}\int_{0}^1z(t)^3dt$ to the next line? $\endgroup$
    – EgoKilla
    Commented Nov 12, 2019 at 3:34
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    $\begingroup$ @EgoKilla I used integration by parts at that step. $\endgroup$ Commented Nov 12, 2019 at 3:35
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    $\begingroup$ @EgoKilla I think that I wrote them in function form might be a bit confusing but it's basically $\int z^3 dt = z^3 t - \int t d(z^3)$. $\endgroup$ Commented Nov 12, 2019 at 3:46
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Let's do this integral in a fun way that most people don't think about. I am going to turn the parabola upside down and consider the volume between the paraboloid and the $z=4$ plane for a very odd reason I will talk about later. In spherical coordinates we have that

$$z=x^2+y^2 \implies \rho^2\cos^2\phi + \rho\cos\phi -\rho^2 = 0$$

and we will solve for $\cos\phi$ instead. We get that

$$\cos\phi = \frac{1}{2\rho}\left(-1\pm\sqrt{1+4\rho^2}\right)$$

we can choose the correct sign by observing that

$$z = -\frac{1}{2}\pm \frac{\sqrt{1+4\rho^2}}{2}$$

In other words, the sign of the square root we choose depends on the sign of $z+\frac{1}{2}$ (this is why I chose to turn the paraboloid upside down, because the parabola in that case switched signs at $z=\frac{1}{2}$ which is was in the middle of our domain, and would've necessitated splitting up the integral into even further pieces). Fortunately, our $z$ are always positive so we only have to take the positive root.

We can write the integral of this paraboloid into two separate integrals doing the $d\phi$ first instead of $d\rho$.

$$ I = \int_0^{2\pi} \int_0^4 \int_0^{\cos^{-1}\left(\frac{-1+\sqrt{1+4\rho^2}}{2\rho}\right)} \rho^2 \sin\phi d\phi d\rho d\theta + \int_0^{2\pi} \int_4^{\sqrt{20}} \int_{\cos^{-1}\left(\frac{4}{\rho} \right)}^{\cos^{-1}\left(\frac{-1+\sqrt{1+4\rho^2}}{2\rho}\right)} \rho^2 \sin\phi d\phi d\rho d\theta$$

where we used the fact that for the planar boundary $z=4 \implies \rho = 4\sec\phi$.

Why is this nicer? Because our integrand evaluates to a $\cos\phi$ and cancels out with the $\cos^{-1}$ in the bounds when they are plugged in. Continuing, we get

$$= 2\pi \Biggr[ \int_0^4 \rho^2 + \frac{1}{2}\rho - \frac{1}{2}\rho\sqrt{1+4\rho^2}d\rho + \int_4^{\sqrt{20}} 4\rho + \frac{1}{2}\rho - \frac{1}{2}\rho\sqrt{1+4\rho^2}d\rho \Biggr]$$

$$= 2\pi\Biggr[\int_0^4 \rho^2 d\rho + \int_0^{\sqrt{20}} \frac{1}{2}\rho - \frac{1}{2}\rho\sqrt{1+4\rho^2}d\rho + \int_4^{\sqrt{20}} 4\rho d\rho \Biggr]$$

$$= 2\pi \Biggr[ \frac{64}{3} + 5 - \frac{1}{24}(1+4\rho^2)^{\frac{3}{2}}\Bigr|_0^{\sqrt{20}} + 8 \Biggr] = 2\pi \Biggr[\frac{103}{3} - \frac{9^3-1}{24}\Biggr] = \boxed{8\pi}$$

The beauty of this method is that there is nothing more computationally complex than the integral of a few polynomial like terms. But the fastest way to do this integral was in cylindrical coordinates with $dr$ first. Perhaps, though, there might be a situation where the integrand density may have nice spherical form over its cylindrical representation.

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