4
$\begingroup$

We are told to evaluate the triple integral:

$$\iiint_E z dV$$

where $E$ is bounded by $x=4y^2+4z^2$ and $x=4$.

My attempt:

First I noticed that this represents a paraboloid on the x axis so I thought to use cylindrical coordinates (however as the paraboloid was centered around x I wasnt sure whether to let $z=r\cos\theta$, $y=r\sin\theta$ or the other way around?) $$z=r\cos\theta$$ $$y=r\sin\theta$$ $$0<r<\frac{\sqrt x}{2}$$ $$0<\theta<2\pi$$ $$0<x<4$$ and our integral becomes $$\int_0^4\int_0^{2\pi}\int_0^{\frac{\sqrt x}{2}} r\cos\theta r dr d\theta dx=0$$

However my textbook James Stewart Calculus gives me $\frac{16\pi}{3}$.

Where have I gone wrong?

$\endgroup$
1
  • 1
    $\begingroup$ $\iiint_E z\,dV=0$ sounds right to me because of the symmetry of $E$. Are you sure the integral in Stewart isn't $\iint_E x\,dV$? $\endgroup$ Commented Jan 4, 2023 at 15:10

1 Answer 1

5
$\begingroup$

your answer for the integral $$\iiint_E z dV = 0$$ is correct since it is followed by the symmetry of $z$ in relation to the axis $x$.

instead, if you evaluate the integral: $$\iiint_E x dV$$ you will get: \begin{align*} \int_0^4\int_0^{2\pi}\int_0^{\frac{\sqrt x}{2}} x r \,dr \,d\theta \,dx &= \int_0^4\int_0^{2\pi}x^2\cdot\left(\frac{1}{8}\right)\,d\theta\,dx \\&=\int_0^4x^2\cdot\left(\frac{1}{8}\right)\cdot(2\pi)\, dx \\&=\frac{\pi}{4} \int_0^4x^2\,dx = \frac{16\pi}{3} \end{align*}

$\endgroup$
4
  • 1
    $\begingroup$ Good answer. I cleaned up your LaTeX a bit; hope you don't mind. $\endgroup$ Commented Jan 4, 2023 at 15:38
  • $\begingroup$ Thanks, it must be an error in the textbook. In a situation like this how do you know whether $z=r\cos\theta$ or $z=r\sin\theta$ ? $\endgroup$
    – zak zaki
    Commented Jan 4, 2023 at 15:38
  • $\begingroup$ @zakzaki In order to preserve orientation you would want $y=r \cos \theta$ and $z = r \sin \theta$. Or you could just swap the $z$ and $x$ coordinates and use the usual $x=r\cos\theta$ and $y=r \sin\theta$ cylindrical coordinates. I don't think it matters here. $\endgroup$ Commented Jan 4, 2023 at 15:41
  • $\begingroup$ assigning the coordinates such as z=rcosθ or z=rsinθ is arbitrary and is by choice, you should be able to evaluate the same values by both of them but you must be careful when you are deriving your dV and assigning boundary values to your z, here both of them preserve the symmetry of z in relation to x. $\endgroup$ Commented Jan 4, 2023 at 15:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .