Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the xy-plane

Question

Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the xy-plane

in progress 0
Verity 5 years 2021-08-10T03:54:39+00:00 1 Answers 15 views 0

Answers ( )

  1. taiduc
    0
    2021-08-10T03:56:18+00:00
    Reply

    Answer:

    The area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

    Step-by-step explanation:

    The surface area of the sphere is:

    \int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA

    and the cylinder x^2 + y^2 =ax can be written as:

    r^2 = arcos \theta

    r = a cos \theta

    where;

    D = domain of integration which spans between \{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}

    and;

    the part of the sphere:

    x^2 + y^2 + z^2 = a^2

    making z the subject of the formula, then :

    z = \sqrt{a^2 - (x^2 +y^2)}

    Thus,

    \dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}

    \dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}

    Similarly;

    \dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}

    \dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}

    So;

    \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}[/tex]

    \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}

    \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}

    \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}

    From cylindrical coordinates; we have:

    \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}

    dA = rdrdθ

    By applying the symmetry in the x-axis, the area of the surface will be:

    A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA

    A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta

    A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta

    A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta[tex]A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta[/tex]

    A = 2a^2 [ cos \theta + \theta ]^{\dfrac{\pi}{2} }_{0}

    A = 2a^2 [ cos \dfrac{\pi}{2}+ \dfrac{\pi}{2} - cos (0)- (0)]

    A = 2a^2 [0 + \dfrac{\pi}{2}-1+0]

    A = a^2 \pi - 2a^2

    \mathbf{A = a^2 ( \pi -2)}

    Therefore, the area of the sphere in the cylinder and which locate above the xy plane is \mathbf{ a^2 ( \pi -2)}

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )