Calc 3 triple integral bounds

Question

I was given a region below a hemisphere of $z=\sqrt{25-x^2-y^2}$ and $z=3$ in the order of $dr\cdot dz \cdot d \theta$ for Cartesian and $dp\cdot d \phi\cdot d \theta$ for spherical. I tried finding the bounds for $r$ and I had $0$ to $\sqrt{25-z^2}$ but I asked a friend and he told me that we were supposed to set the $z$ equal to each other and solve for the radius or should I do that for the spherical coordinates only? Sorry if it’s confusing I’ll try to describe further if you have any questions.

Answer 1

So the bounds we have are: $$3\le z\le \sqrt {25-x^2-y^2} $$ Now, considering the shape of our graph (i.e. Dome-shaped) we can use cylindrical coordinate. As you know: $$x^2+y^2=r^2 \implies 3 \le\ z\le \sqrt{25-r^2}$$ Since the graph $\sqrt{25-x^2-y^2}$ projects a circle on the $x-y$ axis, we can deduce $0\le \theta \le 2\pi $.

Also, Notice $r$ is maximum when $z=3$, because that is when we have the largest circle projected on $x-y$ axis. Now, to get the upper bound for $r$ we must set $z=3$:$$3=\sqrt{25-r^2}$$$$9=25-r^2$$$$r^2=16$$$$r=4$$ Therefore, $0\le r\le 4$.

Keeping those in mind, our integral then becomes $$\int_{0}^{2\pi}\int_0^4\int_3^{\sqrt{25-r^2}}rdzdrd\theta$$

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$(r\rightarrow z )\land(z\rightarrow r)$

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$$\int_{0}^{2\pi}\int_0^4\int_3^{\sqrt{25-z^2}}zdrdzd\theta$$

Answer 2

similar answer

Found a similar answer to this that makes it possible to be integrated that way