Multivariable Calculus – Triple Integral Above Paraboloid and Below Plane

multivariable-calculus

$\bf\sf37.$ Evaluate $\iiint_E z\,dV,$ where $E$ lies above the paraboloid $z=x^2+y^2$ and below the plane $z=2y.$

In cylindrical coordinates the paraboloid is given by $z=r^2$ and the plane by $z=2r\sin\theta$ and they intersect in the circle $r=2\sin\theta$. Then $\iiint_E z\,dV=\int_0^\pi\int_0^{2r\sin\theta}rz\,dz\,dr\,d\theta=\tfrac{5\pi}6.$ [using a CAS]

The solution claims that the domain of integration $= \{ \, 0 \le \theta \le \color{red}{ \pi }, 0 \le r \le 2\sin\theta, r^2 \le z \le 2r \sin\theta \, \} $. Why isn’t the upper bound on $\theta$: $\color{red}{ 2 \pi } $?

$1.$ I know $ x^2 + y^2 = 2y \iff r = 2\sin\theta$. But how does this become the upper bound on $r$?

$2.$ Conforming to the solution, the projection (onto $z=0$) of the intersection of paraboloid and plane $= x^2 + y^2 = 2y \iff x^2 + (y – 1)^2 = 1$, a circle of radius 1 centered at $(0, 1)$.
Aren’t we concerned with this entire circle?

I tried to plot this: https://mathematica.stackexchange.com/q/45059/9983

Best Answer

  1. The projection is a circle as you indicated.
  2. The circle passing through the origin.
  3. The circle has center at $\color{red}{(0,1)}$ not $(0,0.5)$ and it has radius $\color{red}{1}$

So, the circle lies in the upper half of the xy-plane where $y\ge 0$.

  1. The radial arm ranges from o to 1
  2. The radial arm starts at $\theta=0$ on the positive x axis
  3. The radial arm covers the disk at $\theta=\pi$ on the negative x axis.

The following figures represent the projection on xy-plane

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