# triple integral bounded by cylinder

Evaluate the triple integral where E is the solid bounded by the cylinder and the planes and in the first octant.

I tried really hard to get the write answer but I am still struggling in the picture so I could not find the limits

you can check if this is correct: $$\int_{\theta=0}^{\pi/2}\int_{r=0}^3\int_{x=0}^{\frac{rcos\theta}{3}}r^2sin\theta dxdrd\theta$$

We have a cylinder with a circular cross-section of radius 3 oriented with its symmetry axis on the $\ x-$ axis (which points off to the left in the graph above). The volume is bounded by the $\ xy - \ \text{and} \ \ yz-$ planes and the oblique plane $\ y = 3x \$ , so it is a narrow wedge or "sliver" in the first octant.

The projection of this volume onto the $\ yz-$ plane is a quarter-cicle, so it is about equally convenient to integrate using either Cartesian or cylindrical coordinates. In the former, since we have the "thickness" of the wedge extending in the positive $\ x-$ direction, we can write

$$\ \int_0^3 \int_0^{\sqrt{9 - y^2}} \int_0^{y/3} \ \ dx \ dz \ dy \ \ = \ \ \int_0^3 \int_0^{\sqrt{9 - y^2}} \ x \ \vert_0^{y/3} \ \ dz \ dy$$

$$= \ \ \frac{1}{3} \ \int_0^3 \int_0^{\sqrt{9 - y^2}} \ y \ \ dz \ dy \ \ = \ \ \frac{1}{3} \ \int_0^3 \ yz \ \vert_0^{\sqrt{9 - y^2}} \ \ dy \ \ = \ \ = \ \ \frac{1}{3} \ \int_0^3 \ y \ \sqrt{9 - y^2} \ \ dy \ \ .$$

Making the substitution $\ u \ = \ 9 - y^2 \ , \ du \ = \ - 2y \ dy \ \ \Rightarrow \ \ y \ dy \ = \ -\frac{1}{2} du \$ , we obtain

$$\rightarrow \ \ \frac{1}{3} \ \int_9^0 \ \sqrt{u} \ \ \left( -\frac{1}{2} du \right) \ \ = \ \ \frac{1}{6} \ \cdot \ \frac{2}{3} u^{3/2} \ \vert_0^9 \ \ = \ \frac{1}{9} \cdot \ 9^{3/2} \ = \ \frac{27}{9} \ = \ 3 \ \ .$$

In cylindrical coordinates, we would write the axial coordinate as $\ x = \frac{1}{3} y = \frac{1}{3} r \cos \theta \$ . The integral is then as shown (with one typo) by ketan, which produces

$$\int_0^{\pi / 2} \int_0^3 \int_0^{(r \cos \theta)/ 3} \ \ dx \ \ r \ dr \ \ d \theta \ \ = \ \ \int_0^{\pi / 2} \int_0^3 \ \frac{1}{3} r \cos \theta \ \ r \ dr \ \ d \theta$$

$$= \ \ \frac{1}{3} \ \int_0^{\pi / 2} \int_0^3 \ r^2 \cos \theta \ \ dr \ d \theta \ \ = \ \ \frac{1}{3} \ \int_0^{\pi / 2} \ \frac{1}{3} r^3 \cos \theta \ \vert_{r=0}^{r=3} \ \ d \theta$$

$$= \ \ \frac{1}{9} \ \int_0^{\pi / 2} \ 3^3 \ \cos \theta \ \ \ d \theta \ \ = \ \ 3 \ \sin \theta \ \vert_0^{\pi / 2} \ = \ 3 \ (1 - 0 ) \ = \ 3 \ \ .$$

We can compare this volume to that of a hemispherical "sliver" of radius 3 , which will be slightly smaller. The "opening angle" of the wedge is given by $\ \tan \phi \ = \ \frac{1}{3} \ \Rightarrow \ \phi \ \approx \ 18.4º$ , so the wedge's volume is

$$\ \frac{18.4º}{360º} \ \cdot \ \frac{2 \pi}{3} \ \cdot \ 3^3 \ \approx \ \frac{1}{20} \ \cdot \ \frac{2 \pi}{3} \ \cdot \ 27 \ = \ \frac{54 \pi}{60} \ \approx \ 2.83 \ \ .$$

So our result appears to be reliable.