Question

Find the volume of the solid bounded by the plane z=0 and the paraboloid z=1-x^2 –y^2

Answer 1

Answer:

π/2

Step-by-step explanation:

z = 1 − x² − y²

We can integrate by first converting to cylindrical coordinates.

z = 1 − r²

At z = 0, r = 1.  So the limits of integration are:

0 ≤ r ≤ 1

0 ≤ θ ≤ 2π

0 ≤ z ≤ 1 − r²

The volume is:

V = ∫ ∫ ∫ r dz dr dθ

V = ∫₀²ᴾ ∫₀¹ ∫₀ᶻ r dz dr dθ

V = ∫₀²ᴾ ∫₀¹ rz |₀ᶻ dr dθ

V = ∫₀²ᴾ ∫₀¹ r (1 − r²) dr dθ

V = ∫₀²ᴾ ∫₀¹ (r − r³) dr dθ

V = ∫₀²ᴾ (½ r² − ¼ r⁴) |₀¹ dθ

V = ∫₀²ᴾ ¼ dθ

V = ¼ θ |₀²ᴾ

V = π/2

Another way we can look at this is by slicing the paraboloid into a stack of thin circular discs.  Each disc has a volume of:

dV = π r² dz

We know that r² = 1 - z, so:

dV = π (1 - z) dz

So the volume is:

V = ∫₀¹ π (1 - z) dz

V = π (z - ½ z²) |₀¹

V = π/2

Answer 2

Following are the calculation for the volume:

Plane $\ z = 0$ and paraboloid bind a solid The xy-plane (the plane $\ z = 0$) on the circle is intersected by the paraboloid $z = 1 -x^2 - y^2 .$

$\to 1 = x^2 + y^2$

The paraboloid is represented by the interior of this circle. The integrating limitations in polar coordinates are as follows:

$\to 0 \leq \theta \leq 2\pi ,0 \leq r \leq 1, 0 \leq z \leq 1-r^2$

The element of differential volume is $r dz dr d\theta$.

Calculating the volume:  

$V = \int^{2 \pi}_{0} \int^{1}_{0} \int^{1- r^2}_{0} r\ dz dr d\theta$

   $= \int^{2 \pi}_{0} \int^{1}_{0} [z]^{1- r^2}_{0} r\ dr d\theta\\\\= \int^{2 \pi}_{0} \int^{1}_{0} r(1- r^2) dr d\theta\\\\= \int^{2 \pi}_{0} \int^{1}_{0} (r- r^3) dr d\theta\\\\= \int^{2 \pi}_{0} [\frac{r^2}{2}- \frac{r^4}{4})^{1}_{0} d\theta\\\\= \int^{2 \pi}_{0} \frac{1}{4} d \theta \\\\= \frac{1}{4} d [\theta]^{2 \pi}_{0} \\\\=\frac{\pi}{2}$

$z=1-x^2-y^2$

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