Finding the volume of a solid regionI'm trying to find the volume of the solid region inside the sphere x 2 + y 2 + z 2 = 4, and the upper nappe of the cone z 2 = 3 x 2 + 3 y 2 (I only have to set up the triple integral itself, not evaluate it).So, I need to do this in cylindrical coordinates. Thus, for the r value, since z 2 + y 2 = 4, r = 2. Thus, 0 ≤ r ≤ 2.Next, I need to determine the value of z. I've considered the fact that y = sin ⁡ ( θ ), and x = cos ⁡ ( θ ). Thus, z 2 = 3 ( cos 2 ⁡ ( θ ) ) + 3 ( sin 2 ⁡ ( θ ) ). This is equivalent to z 2 = 3. Thus, z = 3 . This would be the lower limit, because it came from the equation of the cone. For the upper boundary of z, I'm not completely sure. Any help with this would be greatly appreciated.Finally, for θ, I think it would be 0 to 2 π, since it is the solid region inside a sphere.I'm not extremely confident in my abilities yet, and I don't want to lead myself astray. Any help/corrections would be greatly appreciated!

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Finding the volume of a solid regionI&#039;m trying to find the volume of the solid region inside the sphere       x    2    +      y    2    +      z    2    =  4, and the upper nappe of the cone       z    2    =  3      x    2    +  3      y    2   (I only have to set up the triple integral itself, not evaluate it).So, I need to do this in cylindrical coordinates. Thus, for the r value, since       z    2    +      y    2    =  4,   r  =  2. Thus,   0  ≤  r  ≤  2.Next, I need to determine the value of z. I&#039;ve considered the fact that   y  =  sin  ⁡  (  θ  ), and   x  =  cos  ⁡  (  θ  ). Thus,       z    2    =  3  (      cos    2    ⁡  (  θ  )  )  +  3  (      sin    2    ⁡  (  θ  )  ). This is equivalent to       z    2    =  3. Thus,   z  =      3  . This would be the lower limit, because it came from the equation of the cone. For the upper boundary of z, I&#039;m not completely sure. Any help with this would be greatly appreciated.Finally, for   θ, I think it would be 0 to   2  π, since it is the solid region inside a sphere.I&#039;m not extremely confident in my abilities yet, and I don&#039;t want to lead myself astray. Any help/corrections would be greatly appreciated!

Katelyn Chapman · Accepted Answer

Step 1In the conversion between rectangular and cylindrical coordinates,   x  =  r  cos  ⁡  θ and   y  =  r  sin  ⁡  θ (in your attempt, you dropped the r).Then the equation of the cone becomes      z    2    =  3      r    2        cos    2    ⁡  θ  +  3      r    2        sin    2    ⁡  θ  =  3      r    2    ,and since you&#039;re only interested on the upper half, you can use   z  =      3    r.It is important not to drop the r here because in cylindrical coordinates, r is the radius on the xy-plane, not necessarily the distance of the point from the origin, so in a cone it is not fixed; it varies with the height z. (The equation   z  =      3   actually gives you a plane that is parallel to the xy-plane. An equation where r is fixed, like   r  =  2, corresponds to a right circular cylinder.)Now, the equation for your sphere will be   4  =      z    2    +      x    2    +      z    2    =      z    2    +      r    2        cos    2    ⁡  θ  +      r    2        sin    2    ⁡  θ  =      z    2    +      r    2  Here again you&#039;re only interested in the upper half, so solving for z you get   z  =      4    −          r      2      .Step 2The bounds for z will be these equations (the lower one is the cone, the upper one is the sphere). The fact that they depend on r tells you that you must integrate with respect to z before integrating with respect to r.The bounds you found for   θ are correct.To figure out the bounds for r, you need to find the r at which the cone and the sphere intersect:      3    r  =      4    −          r      2          ⟹    3      r    2    =  4  −      r    2      ⟹    4      r    2    =  4    ⟹    r  =  ±  1We can take the positive value and get the limits for r:    0  ≤  r  ≤  1

Find the volume of the solid region inside the sphere x^2+y^2+z^2=4, and the upper nappe of the cone z^2=3x^2+3y^2.

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