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Find the Volume of the Solid that is enclosed by the Cone and the Sphere

This question aims to find the volume of the solid enclosed by the cone and a sphere by using the method of polar coordinates to find the volume. Cylindrical coordinates extend the two-dimensional coordinates to three-dimensional coordinates.

In a sphere, the distance of the origin $(0,0)$ to the point $P$ is called the radius $r$. By joining the line from the origin to the point $P$, the angle made by this radial line from the $x-axis$ is called the angle theta, represented by $\theta$. Radius $r$ and $\theta$ have some values that can be used in limits for integration.

Expert Answer

The $z-axis$ is projected in a cartesian plane along with the $xy$-plane to form a three-dimensional plane. This plane is represented by $(r, \theta, z)$ in terms of polar coordinates.

To find the limits of $z$, we will take the square root of the double cones. The positive square root represents the top of the cone. The equation of the cone is:

\[z = \sqrt{(x^2 + y^2)}\]

The equation of the sphere is:

\[ x^2 + y^2 + z^2 = 2\]

This equation is derived from the polar coordinates formula, where $x^2 + y^2 = r^2$ when $z = r^2$.

Both of these equations can be represented on the cartesian plane:

Put the value of $r^2$ in place of $z^2$ by using polar coordinates:

\[ x^2 + y^2 + z^2 = 2\]

\[r^2 + z^2 = 2\]

\[z = \sqrt{2- r^2}\]

We will equate both equations to find the value of $r$ when $z$ = $r$ by:

\[z = \sqrt{(x^2 + y^2)}\]

\[z = \sqrt{(r^2)}\]

\[z = r\]

To find $r$:

\[r = \sqrt{2 – r^2}\]

\[2r^2 = 2\]

\[r = 1\]

When we enter from the $z-axis$, we will come across top of the sphere and bottom of the cone. We will integrate from $0$ to $2\pi$ in the spherical region. The limits at those points are:

 

\int_{a}^b\int_{c}^d f(x,y)dxdy$

 

\[\int_{0}^{2\pi}\ \int_{0}^1\ \int_{r}^\sqrt{2-r^2} dzrdrd\theta\]

 

Integrate with respect to $z$ and put limits of $z$

 

\[\int_{0}^{2\pi}\ \int_{0}^1\ r\sqrt{2-r^2} – r^2 drd\theta\]

 

We will separate the integrals to substitute $u$:

 

\[\int_{0}^{2\pi} [\int_{0}^1\ r\sqrt{2-r^2}dr – \int_{0}^1 r^2 dr] d\theta\]

 

\[u = 2 – r^2 , du = -2rdr\]

 

By simplification, we get:

 

\[\int_{0}^{2\pi} [\int_{1}^2 \frac{-1}{2} \sqrt{u}du \ – \int_{0}^1 r^2 dr] d\theta\]

 

\[\int_{0}^{2\pi} [\int_{1}^2 \frac{1}{2} \sqrt{u}du\ – \int_{0}^1 r^2 dr] d\theta\]

 

Integrating with respect to $u$ and $r$:

 

\[\int_{0}^{2\pi} [\int_{1}^2 \frac{1}{2} \sqrt{u}du\ – \int_{0}^1 r^2 dr] d\theta\]

 

\[\int_{0}^{2\pi}\ \frac{2}{3} (\sqrt{2} – 1) d\theta\]

 

Numerical Solution:

Integration with respect to $\theta$ and then putting its limits give us:

\[V = \frac{4\pi}{3} \large(\sqrt{2} – 1)\]

Image/Mathematical drawings are created in Geogebra

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