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Find the volume of the given solid, bounded by the cylinder y^(2)+z^(2)=64 and the planes x=2y,x=0,z=0 in the first octant

Find the volume of the given solid,\newlinebounded by the cylinder y2+z2=64 y^{2}+z^{2}=64 and the planes x=2y,x=0,z=0 x=2 y, x=0, z=0 in the first octant

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Q. Find the volume of the given solid,\newlinebounded by the cylinder y2+z2=64 y^{2}+z^{2}=64 and the planes x=2y,x=0,z=0 x=2 y, x=0, z=0 in the first octant
  1. Set Up Triple Integral: To find the volume of the solid, we need to set up a triple integral in the appropriate coordinates. Since we are dealing with a cylinder, cylindrical coordinates might be a natural choice. However, since the bounds involve xx and yy explicitly, and we are only in the first octant, it might be simpler to use Cartesian coordinates. We will integrate over xx, then yy, then zz.
  2. Bounds for zz: First, we need to determine the bounds for zz. Since we are in the first octant and bounded by the cylinder y2+z2=64y^2 + z^2 = 64, zz will range from 00 to the positive square root of (64−y2)(64 - y^2).
  3. Bounds for yy: Next, we determine the bounds for yy. The solid is bounded by the plane x=2yx = 2y and x=0x = 0. Since x=0x = 0 is the y-axis and we are in the first octant, yy will range from 00 to x2\frac{x}{2}.
  4. Bounds for x: Finally, we determine the bounds for xx. The solid is bounded by the plane x=2yx = 2y, but since we are in the first octant and yy is bounded by the cylinder, we need to find the intersection of x=2yx = 2y with the cylinder y2+z2=64y^2 + z^2 = 64. Substituting x=2yx = 2y into the cylinder equation gives us (x/2)2+z2=64(x/2)^2 + z^2 = 64, which simplifies to x2/4+z2=64x^2/4 + z^2 = 64. Solving for xx when z=0z = 0 gives us x=2yx = 2y00. Therefore, xx will range from x=2yx = 2y22 to x=2yx = 2y33.
  5. Set Up Volume Integral: Now we can set up the triple integral for the volume VV:V=∫∫∫dV,V = \int\int\int dV, where dV=dz dy dxdV = dz\, dy\, dx. The limits of integration will be from z=0z = 0 to z=64−y2z = \sqrt{64 - y^2}, y=0y = 0 to y=x2y = \frac{x}{2}, and x=0x = 0 to x=16x = 16.
  6. Integrate with Respect to z: The triple integral becomes:\newlineV=∫x=0x=16(∫y=0y=x2(∫z=0z=64−y2dz)dy)dx.V = \int_{x=0}^{x=16} \left(\int_{y=0}^{y=\frac{x}{2}} \left(\int_{z=0}^{z=\sqrt{64 - y^2}} dz\right) dy\right) dx.
  7. Integrate with Respect to y: We first integrate with respect to z: ∫z=0z=64−y2dz=[z]064−y2=64−y2\int_{z=0}^{z=\sqrt{64 - y^2}} dz = [z]_{0}^{\sqrt{64 - y^2}} = \sqrt{64 - y^2}.
  8. Trigonometric Substitution: Substituting this into our integral, we get: V=∫x=0x=16(∫y=0y=x264−y2 dy)dxV = \int_{x=0}^{x=16} \left(\int_{y=0}^{y=\frac{x}{2}} \sqrt{64 - y^2} \, dy\right) dx.
  9. Simplify Integral: Next, we integrate with respect to yy. This is a bit more complex, as it involves an integral of a square root function. We can use a trigonometric substitution, y=8sin⁡(θ)y = 8\sin(\theta), dy=8cos⁡(θ)dθdy = 8\cos(\theta)d\theta, and the integral becomes: ∫y=0y=x/264−y2dy=∫θ=0θ=arcsin⁡(x/16)64−64sin⁡2(θ)⋅8cos⁡(θ)dθ.\int_{y=0}^{y=x/2} \sqrt{64 - y^2} dy = \int_{\theta=0}^{\theta=\arcsin(x/16)} \sqrt{64 - 64\sin^2(\theta)} \cdot 8\cos(\theta) d\theta.
  10. Double Angle Formula: Simplifying the integral, we get: ∫θ=0θ=arcsin⁡(x16)864(1−sin⁡2(θ))⋅cos⁡(θ)dθ=∫θ=0θ=arcsin⁡(x16)64cos⁡2(θ)dθ\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 8\sqrt{64(1 - \sin^2(\theta))} \cdot \cos(\theta) d\theta = \int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 64\cos^2(\theta) d\theta.
  11. Evaluate Integral Bounds: This integral can be solved using the double angle formula for cosine, cos⁡2(θ)=1+cos⁡(2θ)2\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}. The integral becomes:\newline∫θ=0θ=arcsin⁡(x16)64⋅1+cos⁡(2θ)2dθ=32∫θ=0θ=arcsin⁡(x16)(1+cos⁡(2θ))dθ\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 64 \cdot \frac{1 + \cos(2\theta)}{2} d\theta = 32\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} (1 + \cos(2\theta)) d\theta.
  12. Substitute Back into Integral: Integrating with respect to θ\theta gives us: 32[θ+(1/2)sin⁡(2θ)]32[\theta + (1/2)\sin(2\theta)] from θ=0\theta=0 to θ=arcsin⁡(x/16)\theta=\arcsin(x/16).
  13. Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: 32[arcsin⁡(x16)+(12)sin⁡(2arcsin⁡(x16))]−32[0+(12)sin⁡(0)]32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))] - 32[0 + (\frac{1}{2})\sin(0)].
  14. Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: \newline32[arcsin⁡(x16)+(12)sin⁡(2arcsin⁡(x16))]−32[0+(12)sin⁡(0)]32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))] - 32[0 + (\frac{1}{2})\sin(0)].Since sin⁡(2arcsin⁡(x16))=2sin⁡(arcsin⁡(x16))cos⁡(arcsin⁡(x16))\sin(2\arcsin(\frac{x}{16})) = 2\sin(\arcsin(\frac{x}{16}))\cos(\arcsin(\frac{x}{16})) and sin⁡(arcsin⁡(x16))=x16\sin(\arcsin(\frac{x}{16})) = \frac{x}{16}, cos⁡(arcsin⁡(x16))=1−(x16)2\cos(\arcsin(\frac{x}{16})) = \sqrt{1 - (\frac{x}{16})^2}, the expression simplifies to: \newline32[arcsin⁡(x16)+(12)(2(x16)1−(x16)2)]32[\arcsin(\frac{x}{16}) + (\frac{1}{2})(2(\frac{x}{16})\sqrt{1 - (\frac{x}{16})^2})].
  15. Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: \newline32[arcsin⁡(x16)+(12)sin⁡(2arcsin⁡(x16))]32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))] - 32[0+(12)sin⁡(0)]32[0 + (\frac{1}{2})\sin(0)].Since sin⁡(2arcsin⁡(x16))=2sin⁡(arcsin⁡(x16))cos⁡(arcsin⁡(x16))\sin(2\arcsin(\frac{x}{16})) = 2\sin(\arcsin(\frac{x}{16}))\cos(\arcsin(\frac{x}{16})) and sin⁡(arcsin⁡(x16))=x16\sin(\arcsin(\frac{x}{16})) = \frac{x}{16}, cos⁡(arcsin⁡(x16))=1−(x16)2\cos(\arcsin(\frac{x}{16})) = \sqrt{1 - (\frac{x}{16})^2}, the expression simplifies to: \newline32[arcsin⁡(x16)+(12)(2(x16)1−(x16)2)]32[\arcsin(\frac{x}{16}) + (\frac{1}{2})(2(\frac{x}{16})\sqrt{1 - (\frac{x}{16})^2})].Now we substitute this expression back into the integral for x and integrate from x=0x=0 to x=16x=16: \newlineV=∫x=0x=1632[arcsin⁡(x16)+(x8)1−(x16)2]dxV = \int_{x=0}^{x=16} 32[\arcsin(\frac{x}{16}) + (\frac{x}{8})\sqrt{1 - (\frac{x}{16})^2}] dx.
  16. Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: 32[arcsin⁡(x/16)+(1/2)sin⁡(2arcsin⁡(x/16))]−32[0+(1/2)sin⁡(0)]32[\arcsin(x/16) + (1/2)\sin(2\arcsin(x/16))] - 32[0 + (1/2)\sin(0)]. Since sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16)) and sin⁡(arcsin⁡(x/16))=x/16\sin(\arcsin(x/16)) = x/16, cos⁡(arcsin⁡(x/16))=1−(x/16)2\cos(\arcsin(x/16)) = \sqrt{1 - (x/16)^2}, the expression simplifies to: 32[arcsin⁡(x/16)+(1/2)(2(x/16)1−(x/16)2)]32[\arcsin(x/16) + (1/2)(2(x/16)\sqrt{1 - (x/16)^2})]. Now we substitute this expression back into the integral for xx and integrate from x=0x=0 to x=16x=16: V=∫x=0x=1632[arcsin⁡(x/16)+(x/8)1−(x/16)2] dxV = \int_{x=0}^{x=16} 32[\arcsin(x/16) + (x/8)\sqrt{1 - (x/16)^2}] \, dx. This integral is quite complex and typically requires numerical methods or special functions to evaluate. However, we have made a mistake in our setup. The upper limit for yy should not be sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))00, but rather the intersection of the plane sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))11 with the cylinder sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))22 when sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))33. This means yy should range from sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))55 to sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))66, not sin⁡(2arcsin⁡(x/16))=2sin⁡(arcsin⁡(x/16))cos⁡(arcsin⁡(x/16))\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))00. We need to correct this error and re-evaluate the integral.

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