zidane
  • zidane
vector calculus find the volume inside the surfaces z = x^2 + y^2 and z = surd (2 - x^2 - y^2) i need to know the steps in order to get the answer.help me.tq
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
zidane
  • zidane
help !!
anonymous
  • anonymous
Still need help?
anonymous
  • anonymous
If you plot your surfaces, you'll find that \[z=\sqrt{2-x^2-y^2}\]will sit on top of \[z=x^2+y^2\] and you're basically calculating the volume between them. If you were to calculate the volume of the first and then the volume of the second, and subtract the second, the remainder would be the volume in between (you should try to plot something to see this happening). Now, we need to work out the domain over which we integrate. The two surfaces intersect on the curve \[x^2+y^2=1\] So we have for the net volume,\[V=\int\limits_{D}^{}\int\limits_{}^{}z_2-z_1dA=V=\int\limits_{D}^{}\int\limits_{}^{}\sqrt{2-x^2-y^2}-(x^2+y^2)dA\]where our domain is \[D=\left\{ (x,y)|x^2+y^2=1 \right\}\]

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anonymous
  • anonymous
We can make life less complicated by switching to polar coordinates, noting the transformation,\[x=r \cos \theta , y= r \sin \theta , dA= r dr d \theta\]and where our domain is (in these coordinates),\[D_{r,\theta}=\left\{ (r,\theta)|0 \le r \le 1, 0 \le \theta \le 2\pi \right\}\]The integral then becomes,\[V=\int\limits_{0}^{1}\int\limits_{0}^{2\pi}(\sqrt{2-r^2}-r^2)r dr d \theta=\int\limits_{0}^{1}\int\limits_{0}^{2\pi}r \sqrt{2-r^2}-r^3 dr d \theta\]
anonymous
  • anonymous
The integral is independent of theta, so you can integrate that out immediately to be left with,\[V=2\pi \int\limits_{0}^{1}r \sqrt{2-r^2}-r^3 d r\]
anonymous
  • anonymous
i.e.\[V=2 \pi \int\limits_{0}^{1}r \sqrt{2-r^2}dr-2 \pi \int\limits_{0}^{1}r^3 dr\]
anonymous
  • anonymous
The integral on the left-hand side can be evaluated using the substitution,\[u=2-r^2 \rightarrow du=-2r dr \rightarrow rdr = -\frac{du}{2}\]and so the LHS integral is,\[V_L=2\pi \int\limits_{u_1}^{u_2}-\frac{du}{2}u^{1/2}= -2 \pi \frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}|_{u_1}^{u_2}=-\frac{2\pi}{3}u^{3/2}|_{u_1}^{u_2}\]
anonymous
  • anonymous
That is\[V_L=-\frac{2\pi}{3}(2-r^2)^{3/2}|_0^1=\frac{2 \pi}{3}(2\sqrt{2}-1)\]
anonymous
  • anonymous
The second component is\[V_R=-2\pi \frac{r^4}{4}|_0^1=-\frac{\pi}{2}\]
anonymous
  • anonymous
So\[V=V_L+V_R=\frac{2\pi}{3}(2\sqrt{2}-1)-\frac{\pi}{2}\]
anonymous
  • anonymous
If I've stuffed up anywhere, it's a calculation error. The method is sound. Just check what surface is on top of the other, their curve of intersection will define the boundary of the domain of integration. The domain is given on the x-y plane as the projection of the object onto the surface (imagine holding a light above the surface and taking the shadow on the floor as being the domain). If you need help, let me know. Fan me if you haven't; this took ages to write out ;p
zidane
  • zidane
thanks.u help me out.thanks my hero =)
zidane
  • zidane
thanks.u help me out.thanks my hero =)
zidane
  • zidane
thanks.u help me out.thanks my hero =)
anonymous
  • anonymous
zidane, just check everything before you take it on board. I rushed through it. E.g. check the orientation of the surfaces, their intersection, projection of domain, etc.
anonymous
  • anonymous
and you're welcome :)
zidane
  • zidane
your answer is correct.but how come u get x^2 + y^2 =1?
anonymous
  • anonymous
Ah, the projection of the curve where z_1 and z_2 intersect is the boundary of the domain. You can imagine the paraboloid opening upward from the origin and the other curve opening downward from a starting point above. They're both expanding as they open and they will intersect at some height, z. Since they're sharing the same height, we can find all (x,y) that both surfaces share. This set of (x,y) forms the boundary of the domain. So, for z_1=z_2, you'd have \[x^2+y^2=\sqrt{2-x^2-y^2}\] which, when solved, gives the unit circle.
zidane
  • zidane
yeah.i get it now.thanks lokisan =)
anonymous
  • anonymous
np :)

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