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

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Q. Find the volume of the given solid,

\newline

bounded by the cylinder

y^{2}+z^{2}=64

and the planes

x=2 y, x=0, z=0

in the first octant

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 $x$ and $y$ explicitly, and we are only in the first octant, it might be simpler to use Cartesian coordinates. We will integrate over $x$ , then $y$ , then $z$ .
Bounds for $z$ : First, we need to determine the bounds for $z$ . Since we are in the first octant and bounded by the cylinder $y^2 + z^2 = 64$ , $z$ will range from $0$ to the positive square root of $(64 - y^2)$ .
Bounds for $y$ : Next, we determine the bounds for $y$ . The solid is bounded by the plane $x = 2y$ and $x = 0$ . Since $x = 0$ is the y-axis and we are in the first octant, $y$ will range from $0$ to $\frac{x}{2}$ .
Bounds for x: Finally, we determine the bounds for $x$ . The solid is bounded by the plane $x = 2y$ , but since we are in the first octant and $y$ is bounded by the cylinder, we need to find the intersection of $x = 2y$ with the cylinder $y^2 + z^2 = 64$ . Substituting $x = 2y$ into the cylinder equation gives us $(x/2)^2 + z^2 = 64$ , which simplifies to $x^2/4 + z^2 = 64$ . Solving for $x$ when $z = 0$ gives us $x = 2y$ $0$ . Therefore, $x$ will range from $x = 2y$ $2$ to $x = 2y$ $3$ .
Set Up Volume Integral: Now we can set up the triple integral for the volume $V$ : $V = \int\int\int dV,$ where $dV = dz\, dy\, dx$ . The limits of integration will be from $z = 0$ to $z = \sqrt{64 - y^2}$ , $y = 0$ to $y = \frac{x}{2}$ , and $x = 0$ to $x = 16$ .
Integrate with Respect to z: The triple integral becomes: $\newline$ $V = \int_{x=0}^{x=16} \left(\int_{y=0}^{y=\frac{x}{2}} \left(\int_{z=0}^{z=\sqrt{64 - y^2}} dz\right) dy\right) dx.$
Integrate with Respect to y: We first integrate with respect to z: $\int_{z=0}^{z=\sqrt{64 - y^2}} dz = [z]_{0}^{\sqrt{64 - y^2}} = \sqrt{64 - y^2}$ .
Trigonometric Substitution: Substituting this into our integral, we get: $V = \int_{x=0}^{x=16} \left(\int_{y=0}^{y=\frac{x}{2}} \sqrt{64 - y^2} \, dy\right) dx$ .
Simplify Integral: Next, we integrate with respect to $y$ . This is a bit more complex, as it involves an integral of a square root function. We can use a trigonometric substitution, $y = 8\sin(\theta)$ , $dy = 8\cos(\theta)d\theta$ , and the integral becomes: $\int_{y=0}^{y=x/2} \sqrt{64 - y^2} dy = \int_{\theta=0}^{\theta=\arcsin(x/16)} \sqrt{64 - 64\sin^2(\theta)} \cdot 8\cos(\theta) d\theta.$
Double Angle Formula: Simplifying the integral, we get: $\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 8\sqrt{64(1 - \sin^2(\theta))} \cdot \cos(\theta) d\theta = \int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 64\cos^2(\theta) d\theta$ .
Evaluate Integral Bounds: This integral can be solved using the double angle formula for cosine, $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$ . The integral becomes: $\newline$ $\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} 64 \cdot \frac{1 + \cos(2\theta)}{2} d\theta = 32\int_{\theta=0}^{\theta=\arcsin(\frac{x}{16})} (1 + \cos(2\theta)) d\theta$ .
Substitute Back into Integral: Integrating with respect to $\theta$ gives us: $32[\theta + (1/2)\sin(2\theta)]$ from $\theta=0$ to $\theta=\arcsin(x/16)$ .
Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: $32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))] - 32[0 + (\frac{1}{2})\sin(0)]$ .
Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: $\newline$ $32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))] - 32[0 + (\frac{1}{2})\sin(0)]$ .Since $\sin(2\arcsin(\frac{x}{16})) = 2\sin(\arcsin(\frac{x}{16}))\cos(\arcsin(\frac{x}{16}))$ and $\sin(\arcsin(\frac{x}{16})) = \frac{x}{16}$ , $\cos(\arcsin(\frac{x}{16})) = \sqrt{1 - (\frac{x}{16})^2}$ , the expression simplifies to: $\newline$ $32[\arcsin(\frac{x}{16}) + (\frac{1}{2})(2(\frac{x}{16})\sqrt{1 - (\frac{x}{16})^2})]$ .
Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: $\newline$ $32[\arcsin(\frac{x}{16}) + (\frac{1}{2})\sin(2\arcsin(\frac{x}{16}))]$ - $32[0 + (\frac{1}{2})\sin(0)]$ .Since $\sin(2\arcsin(\frac{x}{16})) = 2\sin(\arcsin(\frac{x}{16}))\cos(\arcsin(\frac{x}{16}))$ and $\sin(\arcsin(\frac{x}{16})) = \frac{x}{16}$ , $\cos(\arcsin(\frac{x}{16})) = \sqrt{1 - (\frac{x}{16})^2}$ , the expression simplifies to: $\newline$ $32[\arcsin(\frac{x}{16}) + (\frac{1}{2})(2(\frac{x}{16})\sqrt{1 - (\frac{x}{16})^2})]$ .Now we substitute this expression back into the integral for x and integrate from $x=0$ to $x=16$ : $\newline$ $V = \int_{x=0}^{x=16} 32[\arcsin(\frac{x}{16}) + (\frac{x}{8})\sqrt{1 - (\frac{x}{16})^2}] dx$ .
Correct Error and Re-evaluate: Evaluating the integral at the bounds, we get: $32[\arcsin(x/16) + (1/2)\sin(2\arcsin(x/16))] - 32[0 + (1/2)\sin(0)]$ . Since $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ and $\sin(\arcsin(x/16)) = x/16$ , $\cos(\arcsin(x/16)) = \sqrt{1 - (x/16)^2}$ , the expression simplifies to: $32[\arcsin(x/16) + (1/2)(2(x/16)\sqrt{1 - (x/16)^2})]$ . Now we substitute this expression back into the integral for $x$ and integrate from $x=0$ to $x=16$ : $V = \int_{x=0}^{x=16} 32[\arcsin(x/16) + (x/8)\sqrt{1 - (x/16)^2}] \, dx$ . 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 $y$ should not be $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $0$ , but rather the intersection of the plane $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $1$ with the cylinder $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $2$ when $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $3$ . This means $y$ should range from $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $5$ to $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $6$ , not $\sin(2\arcsin(x/16)) = 2\sin(\arcsin(x/16))\cos(\arcsin(x/16))$ $0$ . We need to correct this error and re-evaluate the integral.

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

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