Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 = 72.

Step 1: The intersection curve is only at the origin.

Step 2: Set up the volume integral over the sphere region.

Step 3:Evaluate the integral to find the volume: [tex]\(V = 216\pi\sqrt{2}\).[/tex]

To find the volume of the solid enclosed by the cone [tex]\(z = x^2 + y^2\)[/tex] and the sphere[tex]\(x^2 + y^2 + z^2 = 72\),[/tex] we'll first find the intersection curve of the cone and the sphere in cylindrical coordinates, then set up the triple integral to find the volume.

Step 1: Finding the intersection curve:

The cone equation in cylindrical coordinates becomes [tex]\(z = r^2\)[/tex] and the sphere equation remains the same as [tex]\(r^2 + z^2 = 72\).[/tex]

To find the intersection, we set these two equations equal to each other:

[tex]\[r^2 = r^2 + z^2\][/tex]

Substitute [tex]\(z = r^2\)[/tex] from the cone equation:

[tex]\[r^2 = r^2 + (r^2)^2\][/tex]

[tex]\[r^2 = r^2 + r^4\][/tex]

[tex]\[r^4 = 0\][/tex]

From this, we see that the only solution is (r = 0), which corresponds to the point at the origin.

Step 2: Setting up the integral for volume:

We'll integrate over the region where the cone lies within the sphere, which is the entire sphere. In cylindrical coordinates, the limits of integration are [tex]\(0 \leq r \leq 6\)[/tex] (since the sphere has radius [tex]\(\sqrt{72} = 6\)) and \(0 \leq \theta \leq 2\pi\).[/tex]

The limits for \(z\) are from the cone to the sphere, so it's from [tex]\(r^2\) to \(\sqrt{72-r^2}\).[/tex]

Thus, the volume integral is:

[tex]\[V = \iiint_{E} r \, dz \, dr \, d\theta\][/tex]

Where (E) is the region enclosed by the cone and the sphere.

Step 3: Evaluate the integral:

[tex]\[V = \int_{0}^{2\pi} \int_{0}^{6} \int_{r^2}^{\sqrt{72-r^2}} r \, dz \, dr \, d\theta\][/tex]

Let's evaluate this integral step by step:

[tex]\[V = \int_{0}^{2\pi} \int_{0}^{6} (r\sqrt{72-r^2} - r^3) \, dr \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[-\frac{1}{4}(72-r^2)^{3/2} - \frac{1}{4}r^4\right]_{0}^{6} \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[-\frac{1}{4}(0)^{3/2} - \frac{1}{4}(6^4) - \left(-\frac{1}{4}(72)^{3/2} - \frac{1}{4}(0)^4\right)\right] \, d\theta\][/tex]

[tex]\[V = \int_{0}^{2\pi} \left[\frac{1}{4}(72)^{3/2}\right] \, d\theta\][/tex]

[tex]\[V = \frac{1}{4}(72)^{3/2} \int_{0}^{2\pi} 1 \, d\theta\][/tex]

[tex]\[V = \frac{1}{4}(72)^{3/2} \cdot 2\pi\][/tex]

[tex]\[V = 36\pi \sqrt{72}\][/tex]

[tex]\[V = 36\pi \times 6\sqrt{2}\][/tex]

[tex]\[V = 216\pi\sqrt{2}\][/tex]

So, the volume of the solid enclosed by the cone and the sphere is [tex]\(216\pi\sqrt{2}\).[/tex]

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