Question

Change from rectangular to cylindrical coordinates. (Let r\geq0 and 0\leq\theta\leq2\pi.) a) (-2, 2, 2) b) (-9,9\sqrt{3,6}) c) Use cylindrical coordin

Change from rectangular to cylindrical coordinates. (Let \(r\geq0\) and \(0\leq\theta\leq2\pi\).)
a) \((-2, 2, 2)\)
b) \((-9,9\sqrt{3,6})\)
c) Use cylindrical coordinates.
Evaluate
\(\int\int\int_{E}xdV\)
where E is enclosed by the planes \(z=0\) and
\(z=x+y+10\)
and by the cylinders
\(x^{2}+y^{2}=16\) and \(x^{2}+y^{2}=36\)
d) Use cylindrical coordinates.
Find the volume of the solid that is enclosed by the cone
\(z=\sqrt{x^{2}+y^{2}}\)
and the sphere
\(x^{2}+y^{2}+z^{2}=8\).

Expert Answers (1)

2021-06-10
Step 1
Change from rectangular to cylindrical coordinates.
Let \(r\geq0\) and \(0\leq\theta\leq2\pi\)
a) \((x, y, z)=(-2, 2, 2)\)
Use cylindrical coordinates.
\(r=\sqrt{x^{2}+y^{2}}=\sqrt{(-2)^{3}+2^{3}}=2\sqrt{2}\)
\(\theta=\arctan\left(\frac{y}{x}\right)=\arctan\left(\frac{2}{-2}\right)(-1)=\frac{3\pi}{4}\)
\(z=z=2\)
Therefore, the required coordinates is,
\((r, \theta, z)=\left(2\sqrt{2}, \frac{2\pi}{4}, 2\right)\)
Step 2
b) \((x, y, z)=(-9.9\sqrt{3.6})\)
Use cylindrical coordinates.
\(r=\sqrt{x^{2}+y^{2}}=\sqrt{(-9)^{2}+(9\sqrt{3})}^{2}=18\)
\(\theta=\arctan\left(\frac{y}{x}\right)=\arctan\left(\frac{9\sqrt{3}}{-9}\right)=\arctan(-\sqrt{3})=\frac{2\pi}{3}\)
\(z=z=6\)
Therefore, the required coordinates is,
\((r, \theta, z)=\left(18, \frac{2\pi}{3}, 6\right)\)
Step 3
c) Use cylindrical coordinates, to evaluate \(\int\int_{E}\int xdV\)
Where E is enclosed by the planes \(z=0\) and \(z=x+y+10\) and by the cylinders \(x^{2}+y^{2}=16\) and \(x^{2}+y^{2}=36\)
\(16\leq x^{2}+y^{2}\leq36\)
\(\Rightarrow16\leq r^{2}\leq36\) As \(x^{2}+y^{2}=r^{2}\)
\(\Rightarrow4\leq r\leq6\)
and \(0\leq\theta\leq2\pi\)
and \(0\leq z\leq x+y+10\)
\(\Rightarrow0\leq z\leq r\cos\theta+r\sin\theta+10\) As \(x=r\cos\theta, y=r\sin\theta\)
and \(dV=dxdydz=rdrd\theta dz\)
So, \(\int\int_{E}\int xdV=\int_{0}^{2a}\int_{4}^{6}\int_{0}^{r\cos\theta+r\sin\theta+10}r\cos\theta(rdrd\theta dz)\)
\(=\int_{0}^{2\pi}\int_{4}^{6}\int_{0}^{r\cos\theta+r\sin\theta+10}r^{2}\cos\theta drd\theta dz\)
\(=\int_{0}^{2\pi}\int_{4}^{6}r^{2}(r\cos\theta+r\sin\theta+10)\cos\theta drd\theta\)
\(=\int_{0}^{2\pi}\int_{4}^{6}(r^{3}(\cos^{2}\theta+\sin\theta\cos\theta)+10r^{2}\cos\theta)drd\theta\)
\(=\int_{0}^{2\pi}\left(\frac{1}{4}r^{4}(\cos^{2}\theta+\sin\theta\cos\theta)+\frac{10}{3}r^{3}\cos\theta\right)_{4}^{6}d\theta\)
\(=\int_{0}^{2\pi}\left(\frac{1}{4}(6^{4}-4^{4})(\cos^{2}\theta+\sin\theta\cos\theta)+\frac{10}{3}(6^{3}-4^{3})\cos\theta\right)d\theta\)
\(=\int_{0}^{2\pi}\left(260(\cos^{2}\theta+\sin\theta\cos\theta)+\frac{1520}{3}\cos\theta\right)d\theta\)
\(=260\pi\)
Step 4
d) Use cylindrical coordinates.
Find the volume of the solid that is enclosed by the cone
\(z=\sqrt{x^{2}+y^{2}}\) and the sphere \(x^{2}+y^{2}+z^{2}=8\)
Use cylindrical polar coordinates,
\(x=r\cos\theta, y=r\sin\theta, z=z\)
\(\Rightarrow x^{2}+y^{2}=r^{2}, dV=dxdydz=rdrd\theta dz\)
Limits:
As \(z=\sqrt{x^{2}+y^{2}}\) and \(x^{2}+y^{2}+z^{2}=8\)
\(\Rightarrow2(x^{2}+y^{2})=8\)
\(\Rightarrow x^{2}+y^{2}=4\)
\(r^{2}=4\)
\(\Rightarrow r=2\)
So, \(E=\left\{(r, \theta, z):0\leq r\leq2,\ 0\leq\theta\leq2\pi,\ r\leq z\leq\sqrt{8-r^{2}}\right\}\)
Volume \(=\int\int_{E}\int dxdydz\)
\(=\int_{r=0}^{2}\int_{\theta=0}^{2\pi}\int_{z=r}^{\sqrt{8-r^{2}}}rdrd\theta dz\)
\(=\int_{r=0}^{2}\int_{\theta=0}^{2\pi}r(\sqrt{8-r^{2}}-r)drd\theta\)
\(=\left(\int_{r=0}^{2}(r\sqrt{8-r^{2}}-r^{2})dr\right)\left(\int_{\theta=0}^{2\pi}d\theta\right)\)
\(=\left[\frac{1}{3}(-(8-r^{2})^{3/2}-r^{3}\right]_{0}^{2}[\theta]_{0}^{2\pi}\)
\(=\left(\frac{1}{3}(-(8-4)^{3/2}-8+(8)^{3/2})\right)(2\pi)\)
\(=\left((\sqrt{2}-1)\frac{16}{3}\right)(2\pi)\)
\(=\frac{32(\sqrt{2}-1)\pi}{3}\)
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