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Using a formula from geometry evaluate $\displaystyle \int_0^3 \! \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \sqrt{9-x^2-y^2}\,dy\,dx$.

Attempt. What formula? Any hints please. I know I could use some trigonometric equations for integrals, but I suppose there is a much easier way to evaluate these integrals.

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Are you sure the order of the differentials is correct? –  user17794 May 25 '12 at 7:14
    
@TimDuff Yeah it is correct. –  Koba May 25 '12 at 7:15
    
@Dostre I doubt it - you say you want to integrate $\int_x^x f(x)dx$, or something similar. –  akkkk May 25 '12 at 7:17
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@Dostre Your order of integration should be $y$ first and then $x$ i.e. $\displaystyle \int \int (\cdot) dy dx$ and not $\displaystyle \int \int (\cdot) dx dy$ since the limits of the inner integral are a function of $x$. I have edited the question to reflect this change. –  user17762 May 25 '12 at 7:39
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1 Answer 1

up vote 4 down vote accepted

Essentially you want to integrate $$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} z dy dx$$ where $z = \sqrt{9 - y^2 - x^2}$. What geometric object is this? It might be of help to rewrite it as $$x^2 + y^2 + z^2 = 9$$

Move your mouse over the gray area below if you want further hints and the solution.

The equation $$x^2 + y^2 + z^2 = 9$$ represents a sphere in $3$D. The limits of the integral will give you the portion of the sphere over which you are integrating and hence the integral will give you the volume of the portion of the sphere. In this case, $z$ is always positive. This restricts the object to a hemi-sphere. The $y$ takes negative and positive values. The $x$ takes only positive values which again restricts the object to half of the hemi-sphere. Hence, the portion of the sphere you are integrating is just $1/4$ of the total sphere. The total volume of the sphere is $\dfrac43 \times \pi \times 3^3$. Hence, the integral is $$\dfrac{\dfrac43 \times \pi \times 3^3}{4} = 9 \pi$$

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Oh thats a sphere. So when integrate that $z$ I am treating it as a constant right? –  Koba May 25 '12 at 7:14
    
No. Your $z$ is a function of $x$ and $y$. Also, it is not a complete sphere, the limits will indicate what portion of the sphere it is. –  user17762 May 25 '12 at 7:15
    
I actual do not have to to integrate that $z$. I can just find the volume of this sphere with a radius $3$ and divide it by $2$: $$V=\pi r^3=9\pi$$ So from $0$ to $3$ the volume will be $V=\frac{9\pi}{2}$ –  Koba May 25 '12 at 7:21
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Exactly you do not need to integrate. Though note that we have $z \geq 0$ and $x \geq 0$. Hence, it is $1/4$ the volume of the sphere i.e. $\frac14 \times \frac43 \times \pi \times 3^3 = 9 \pi$ –  user17762 May 25 '12 at 7:32
    
Yeah thanks. Very helpful. And I like the little trick you are using in your answer. I will have to learn it. –  Koba May 25 '12 at 7:36
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