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



Answer to a math 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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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 is dV = r dz dr d\theta .

To ensure we integrate over the entire volume, we need to find the limits of integration:
1. For r , the limits go from 0 to the upper boundary of the ellipsoid.
2. For z , 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 for z 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 to r :

\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} .

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