Question

The cone \(z^{2}=x^{2}+y^{2},\) for \(z \geq 0,\) cuts the sphere \(x^{2}+y^{2}+z^{2}=16\) along a curve \(C\) a. Find the surface area of the sphere below \(C,\) for \(z \geq 0\). b. Find the surface area of the sphere above \(C\). c. Find the surface area of the cone below \(C,\) for \(z \geq 0\).

Short answer

Question: Find the surface areas of three parts formed by the intersection of a cone with equation \(z^2 = x^2 + y^2\) and a sphere with equation \(x^2 + y^2 + z^2 = 16\). The three parts are: a) the sphere below curve \(C\) and above the \(xy\)-plane, b) the sphere above curve \(C\), and c) the cone below curve \(C\) and above the \(xy\)-plane. Answer: a) Surface area of the sphere below curve \(C\): \(16\pi(\sqrt{2} - 1)\) b) Surface area of the sphere above curve \(C\): \(16\pi(3 - \sqrt{2})\) c) Surface area of the cone below curve \(C\): \(8\pi\)

Step-by-step solution

Find the intersection between the cone and the sphere

To find the intersection, we will substitute the expression for \(z^2\) from the cone equation into the sphere equation: $$x^2 + y^2 + (x^2 + y^2) = 16$$ Simplifying this expression, we get: $$2x^2 + 2y^2 = 16$$ Divide both sides by 2: $$x^2 + y^2 = 8$$ This is the equation of a circle with radius \(\sqrt{8}\) in the \(xy\)-plane. So, curve \(C\) is the projection of this circle onto the cone and the sphere.

Parametrize the sphere and the cone

We use spherical coordinates to parametrize the sphere: $$x = 4\sin\phi\cos\theta$$ $$y = 4\sin\phi\sin\theta$$ $$z = 4\cos\phi$$ Where \(0 \le \phi \le \pi\) and \(0 \le \theta \le 2\pi\). To find \(\phi\) corresponding to the intersection, we substitute \(x^2+y^2 = 8\), getting $$8 = 16\sin^2\phi\cos^2\theta + 16\sin^2\phi\sin^2\theta = 16\sin^2\phi(\cos^2\theta+\sin^2\theta) = 16\sin^2\phi$$ Thus, \(\sin^2\phi = \frac{1}{2}\), or \(\sin\phi=\frac{1}{\sqrt{2}}\). Therefore, \(\phi = \frac{\pi}{4}\). The cone can be parametrized in cylindrical coordinates: $$x = r\cos\theta$$ $$y = r\sin\theta$$ $$z = r$$ Where \(0 \le r \le \sqrt{8}\) and \(0 \le \theta \le 2\pi\).

Find the surface area below and above curve \(C\)

Using the parametrization of the sphere, let's first find the surface area below curve \(C\) and above the \(xy\)-plane. Since \(\phi = \frac{\pi}{4}\) at the intersection, we integrate over the range \(0 \le \phi \le \frac{\pi}{4}\): $$A_{\text{sphere below}} = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2\pi} (4^2 \sin\phi) \, d\theta \, d\phi = 16\int_{0}^{\frac{\pi}{4}} \sin\phi \, d\phi \int_{0}^{2\pi} \, d\theta$$ $$A_{\text{sphere below}} = 16(-\cos\phi\big|_{0}^{\frac{\pi}{4}})(\theta\big|_{0}^{2\pi}) = 16\left(-\frac{1}{\sqrt{2}} + 1\right) 2\pi = 16\pi(\sqrt{2} - 1)$$ Now, let's find the surface area of the sphere above curve \(C\). The only difference is the range of integration for \(\phi\): \(\frac{\pi}{4} \le \phi \le \pi\). Thus, we have: $$A_{\text{sphere above}} = \int_{\frac{\pi}{4}}^{\pi} \int_{0}^{2\pi} (4^2 \sin\phi) \,d\theta \, d\phi = 16\int_{\frac{\pi}{4}}^{\pi} \sin\phi \, d\phi \int_{0}^{2\pi} \, d\theta$$ $$A_{\text{sphere above}} = 16(-\cos\phi\big|_{\frac{\pi}{4}}^{\pi})(\theta\big|_{0}^{2\pi}) = 16\left(2 - \frac{1}{\sqrt{2}}\right) 2\pi = 16\pi(3 - \sqrt{2})$$

Find the surface area of the cone below curve \(C\)

Using the parametrization of the cone, we can write the surface area integral as $$A_{\text{cone}} = \int_{0}^{\sqrt{8}} \int_{0}^{2\pi} r \, d\theta \, dr = \int_{0}^{\sqrt{8}} 2\pi r \, dr$$ $$A_{\text{cone}} = 2\pi (r^2/2\big|_{0}^{\sqrt{8}}) = 8\pi$$ Finally, the surface areas of the three parts are: a. Surface area of the sphere below curve \(C\): \(16\pi(\sqrt{2} - 1)\) b. Surface area of the sphere above curve \(C\): \(16\pi(3 - \sqrt{2})\) c. Surface area of the cone below curve \(C\): \(8\pi\)

Key concepts

Sphere and Cone Intersection

The intersection of a sphere and a cone can create intriguing geometrical patterns. In our problem, the cone's equation is given by \( z^2 = x^2 + y^2 \), and the sphere's equation is \( x^2 + y^2 + z^2 = 16 \). To find how these shapes intersect, we substitute the cone's expression for \( z^2 \) into the sphere's equation:\[ x^2 + y^2 + (x^2 + y^2) = 16 \]Simplifying this gives \( 2x^2 + 2y^2 = 16 \). Dividing by 2 yields \( x^2 + y^2 = 8 \), which represents a circle in the \( xy \)-plane with a radius of \( \sqrt{8} \). This circle represents the curve \( C \), where the cone intersects the sphere, projecting an interesting boundary line also known as an intersection curve.

Parametrization in Calculus

Parametrization is a crucial tool in calculus, allowing us to describe geometric shapes and curves in a mathematical framework. Here, we use spherical coordinates to parametrize the sphere. Spherical coordinates are expressed as:

• \( x = 4\sin\phi\cos\theta \)
• \( y = 4\sin\phi\sin\theta \)
• \( z = 4\cos\phi \)

where \( 0 \le \phi \le \pi \) and \( 0 \le \theta \le 2\pi \). These coordinates help delineate the sphere in three-dimensional space. For the cone, cylindrical coordinates make the problem more straightforward:

• \( x = r\cos\theta \)
• \( y = r\sin\theta \)
• \( z = r \)

Here, we use \( 0 \le r \le \sqrt{8} \) and \( 0 \le \theta \le 2\pi \), helping visualize the structure and calculations of the cone.

Integral Calculus for Surface Area

Integral calculus helps compute precise measurements for complex surfaces, such as spheres and cones. To find the surface area of the sphere below the curve \( C \), we integrate over spherical coordinates. With \( \phi \) aligning the intersection at \( \phi = \frac{\pi}{4} \), we identify the portion under curve \( C \) with:\[ A_{\text{sphere below}} = \int_{0}^{\frac{\pi}{4}} \int_{0}^{2\pi} (4^2 \sin\phi) \, d\theta \, d\phi \]The integral evaluates to \( 16\pi(\sqrt{2} - 1) \).
This same process, using \( \phi \) in the range \( \frac{\pi}{4} \le \phi \le \pi \) yields the area above \( C \):\[ A_{\text{sphere above}} = 16\pi(3 - \sqrt{2}) \]For the cone, we parametrize it in cylindrical coordinates to address the surface below \( C \), using integral calculus:\[ A_{\text{cone}} = \int_{0}^{\sqrt{8}} \int_{0}^{2\pi} r \, d\theta \, dr = 8\pi \]These integrals illustrate how calculus beautifully determines the exact areas of surfaces formed by geometric intersections.