Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

olume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. In each case, let the center of the solid ball be the origin and let the plane be zequals4. Then (d) find the volume by evaluating one of the three triple integrals.

Mathematics

1 answer:

PIT_PIT [12.4K] · Answer 1 · 2026-03-10T14:25:03+03:00

Answer:

Step-by-step explanation:

The equation representing the sphere, which has its center at the origin, can be written as $x^2+y^2+z^2 = 64$ . For z equal to 4, we find

$x^2+y^2= 64-16 = 48$ .

This results in a circle with a radius of $4\sqrt[]{3}$ in the x-y plane.

c) We will build on the analysis from earlier to set limits in both Cartesian and polar coordinates. Initially, we recognize that x spans from $-4\sqrt[]{3}$ to $4\sqrt[]{3}$ . This determination is made by fixing y = 0 and identifying the extreme x values that fall on the circle. For y, we observe that it ranges between $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , which holds because y must reside within the interior of the identified circle. Lastly, z will extend from 4 up to the sphere; hence, it varies from 4 to $\sqrt[]{64-x^2-y^2}$ .

The respective triple integral representing the volume of D in Cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Remember that the cylindrical coordinates are expressed as $x=r\cos \theta, y = r\sin \theta,z = z$ , where r denotes the radial distance from the origin projected onto the x-y plane. Also note that $x^2+y^2 = r^2$ . We will derive new limits for each of the transformed coordinates. Recall that due to the prior circular constraint, $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ varies between 0 and $2\pi$ . Furthermore, r starts from the origin and extends to the edge of the circle, with r reaching a maximum of 4\sqrt[]{3}. Lastly, Z increases from the plane z=4 up to the sphere, where it is constrained by \sqrt[]{64-r^2}. Thus, the integral that computes the desired volume is as follows:

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . It’s important to note that the r factor arises from the Jacobian associated with the transition from Cartesian to polar coordinates, ensuring the integral maintains its value. (Explaining how to calculate the Jacobian exceeds the scope of this response).

a) When dealing with spherical coordinates, keep in mind that $z = \rho \cos \phi, y = \rho \sin \phi \sin \theta, x = \rho \sin \phi \cos \theta$ , where $\phi$ denotes the angle formed between the vector and the z axis, varying from 0 to pi. It is crucial to recognize that at z=4, this angle remains constant along the circle we previously identified. Let’s determine the angle by selecting a point on the circle and employing the angle formula between two vectors. Setting z=4 and x=0 gives us y=4 $\sqrt[]{3}$ by taking the positive square root of 48. We will now compute the angle between the vector $a=(0,4\sqrt[]{3},4)$ and vector b =(0,0,1), which represents the unit vector along the z axis. We apply the following formula

$\cos \phi = \frac{a\cdot b}{||a||||b||} = \frac{(0,4\sqrt[]{3},4)\cdot (0,0,1)}{8}= \frac{1}{2}$

Consequently, across the circle, $\phi = \frac{\pi}{3}$ . Observe that rho transitions from the plane z=4 to the sphere, with rho reaching up to 8. Given $z = \rho \cos \phi$ , we have that $\rho = \frac{4}{\cos \phi}$ at the plane. Thus, the corresponding integral is

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{\frac{4}{\cos \phi}}^{8}\rho^2 \sin \phi d\rho d\phi d\theta$ , where the new factor incorporates the Jacobian for the spherical coordinate system.

d) Let’s work with the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$ .

It’s important to observe that the integral can be separated since the inner part remains independent of theta. By implementing the substitution $u = 64-r^2$ , we achieve $\frac{-du}{2} = r dr$ , leading to

$=-2\pi \cdot \left.(\frac{1}{3}(64-r^2)^{\frac{3}{2}}+2r^{2})\right|_0^{4\sqrt[]{3}}=\frac{320\pi}{3}$