MHB How to find volume of cone+hemisphere on the cone using spherical coordinate

[Another1]

View attachment 7767

i have a problem about find volume of hemisphere I do not know the true extent of radius r (0 to ?)

i think...
cone ( 0 < r < R cosec(\theta) )
hemisphere (0 < r < R)

 

[I like Serena]

i have a problem about find volume of hemisphere I do not know the true extent of radius r (0 to ?)

i think...
cone ( 0 < r < R cosec(\theta) )
hemisphere (0 < r < R)

Hi Another, welcome to MHB! (Wave)

Let's start with the cone.
\begin{tikzpicture}
\def\R{4}
\def\h{\R * cot(30)}
\def\Theta{20}
\coordinate (A) at (0,0);
\coordinate (B) at (0,{\h});
\coordinate (C) at ({\R},{\h});
\coordinate (D) at ({\h*tan(\Theta)},{\h});
\draw (A) -- node[above left] {$h$} (B) -- node[above] {R} (C) -- cycle;
\draw[blue, thick] (A) -- node[above left] {$r_{max}$} (D);
\path (A) node at ({90-\Theta/2}:1.3) {$\theta$};
\draw[thick] (A) +(0,1) arc (90:{90-\Theta}:1);
\end{tikzpicture}

We have $\tan 30^\circ = \frac Rh$ and $\cos\theta = \frac h{r_{max}}$ don't we?
Doesn't that mean:
$$0 < r < r_{max} = R\cot 30^\circ \sec\theta$$As for the hemisphere, suppose we translate it to the origin first before calculating its volume.
Then we just have $0 < r < R$ don't we? (Wondering)

 

[Prove It]

i have a problem about find volume of hemisphere I do not know the true extent of radius r (0 to ?)

i think...
cone ( 0 < r < R cosec(\theta) )
hemisphere (0 < r < R)

No calculus needed...

If the radius of the cone and hemisphere is R units, then we can see the height of the cone we can evaluate by

$\displaystyle \begin{align*} \tan{ \left( \theta \right) } &= \frac{O}{A} \\ \tan{ \left( 30^{\circ} \right) } &= \frac{R}{h} \\ \frac{1}{\sqrt{3}} &= \frac{R}{h} \\ \frac{h}{\sqrt{3}} &= R \\ h &= \sqrt{3}\,R \end{align*}$

Volume of the cone:

$\displaystyle \begin{align*} V &= \frac{1}{3}\,\pi\,r^2\,h \\ &= \frac{\pi\,R^2 \left( \sqrt{3}\,R \right) }{3} \\ &= \frac{\sqrt{3}\,\pi\,R^3}{3} \end{align*}$

Volume of the hemisphere:

$\displaystyle \begin{align*} V &= \frac{2}{3}\,\pi\,r^3 \\ &= \frac{2\,\pi\,R^3}{3} \end{align*}$

Thus the total volume is

$\displaystyle \begin{align*} V &= \frac{\sqrt{3}\,\pi\,R^3}{3} + \frac{2\,\pi\,R^3}{3} \\ &= \frac{\left( 2 + \sqrt{3} \right) \pi \, R^3}{3}\,\textrm{ units}^3 \end{align*}$

 

[Another1]

No calculus needed...

If the radius of the cone and hemisphere is R units, then we can see the height of the cone we can evaluate by

$\displaystyle \begin{align*} \tan{ \left( \theta \right) } &= \frac{O}{A} \\ \tan{ \left( 30^{\circ} \right) } &= \frac{R}{h} \\ \frac{1}{\sqrt{3}} &= \frac{R}{h} \\ \frac{h}{\sqrt{3}} &= R \\ h &= \sqrt{3}\,R \end{align*}$

Volume of the cone:

$\displaystyle \begin{align*} V &= \frac{1}{3}\,\pi\,r^2\,h \\ &= \frac{\pi\,R^2 \left( \sqrt{3}\,R \right) }{3} \\ &= \frac{\sqrt{3}\,\pi\,R^3}{3} \end{align*}$

Volume of the hemisphere:

$\displaystyle \begin{align*} V &= \frac{2}{3}\,\pi\,r^3 \\ &= \frac{2\,\pi\,R^3}{3} \end{align*}$

Thus the total volume is

$\displaystyle \begin{align*} V &= \frac{\sqrt{3}\,\pi\,R^3}{3} + \frac{2\,\pi\,R^3}{3} \\ &= \frac{\left( 2 + \sqrt{3} \right) \pi \, R^3}{3}\,\textrm{ units}^3 \end{align*}$

thank

But for this problem, if the surface equation is determined with [cos(theta)]^2 ? How can we find the volume of hemisphere + cone ?.

thankkkkkk

 

[Prove It]

thank

But for this problem, if the surface equation is determined with [cos(theta)]^2 ? How can we find the volume of hemisphere + cone ?.

thankkkkkk

I have no idea what you mean by the surface equation, and the surface itself (an area) has absolutely nothing to do with the space inside it (the volume)...