Question

Use the vector calculus rules: Find the volume of the solid limited by the ellipsoid z^2 + 9r^2 = 9

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Frederik

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To find the volume of the solid limited by the ellipsoid \frac{z^2}{9} + r^2 = 1 , we will use a triple integral in cylindrical coordinates.

The equation of the ellipsoid can be rewritten as\frac{z^2}{3^2} + r^2 = 1 .

In cylindrical coordinates, the volume element isdV = r dz dr d\theta .

To ensure we integrate over the entire volume, we need to find the limits of integration:

1. Forr , the limits go from 0 to the upper boundary of the ellipsoid.

2. Forz , the limits go from the lower half of the ellipsoid to the upper half, which requires solving for the z bounds of integration.

The lower bound forz is 0, while the upper bound can be found by solving the equation of the ellipsoid for z :

\frac{z^2}{3^2} + r^2 = 1

z^2 = 9 - 9r^2

z = \sqrt{9 - 9r^2}

Thus, the volume integral is:

\text{Volume } = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{9-9r^2}} r dz dr d\theta

Now, we evaluate this triple integral to find the volume of the solid.

\begin{aligned} \text{Volume } &= \int_{0}^{2\pi} \int_{0}^{1} \left[ \left. rz \right|_{0}^{\sqrt{9-9r^2}} \right] dr d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{1} r\sqrt{9-9r^2} dr d\theta \end{aligned}

Now we integrate with respect tor :

\int r\sqrt{9-9r^2}dr=-\frac{}{}(1-r^2)^{\frac{3}{2}}\Big|_0^1=1 by integrating 1 with respective to thita from 0 to 2 pie we get volume is 2 pie

Therefore, the \textbf{Volume } of the solid limited by the ellipsoid\frac{z^2}{9} + r^2 = 1 is \boxed{2\pi} .

The equation of the ellipsoid can be rewritten as

In cylindrical coordinates, the volume element is

To ensure we integrate over the entire volume, we need to find the limits of integration:

1. For

2. For

The lower bound for

Thus, the volume integral is:

Now, we evaluate this triple integral to find the volume of the solid.

Now we integrate with respect to

Therefore, the \textbf{Volume } of the solid limited by the ellipsoid

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