Question

Find the volume of the region enclosed by the cylinder x squared plus y squared equals 36 and the planes z equals 0 and y plus z equals 36.

Answer 1

Answer:

$\mathbf{V = 1296 \pi }$

Step-by-step explanation:

Given that :

Find the volume of the region enclosed by the cylinder $x^2 + y^2 =36$ and the plane z = 0 and y + z = 36

From y + z = 36

z = 36 - y

The volume of the region can be represented by the equation:

$V = \int\limits \int\limits_D(36-y)dA$

In this case;

D is the region given by $x^2 + y^2 = 36$

Relating this to polar coordinates

x = rcosθ    y = rsinθ

x² + y² = r²

x² + y² = 36

r² = 36

r = $\sqrt{36}$

r = 6

dA = rdrdθ

r → 0 to 6

θ to 0  to 2π

Therefore:

$V = \int\limits^{2 \pi} _0 \int\limits ^6_0 (36-r sin \theta ) (rdrd \theta)$

$V = \int\limits^{2 \pi} _0 \int\limits ^6_0 (36-r^2 sin \theta ) drd \theta$

$V = \int\limits^{2 \pi} _0 [\dfrac{36r^2}{2}- \dfrac{r^3}{3}sin \theta]^6_0 \ d\theta$

$V = \int\limits^{2 \pi} _0 [648- \dfrac{216}{3}sin \theta]d\theta$

$V = \int\limits^{2 \pi} _0 [648+\dfrac{216}{3}cos \theta]d\theta$

$V = [648+\dfrac{216}{3}cos \theta]^{2 \pi}_0$

$V = [648(2 \pi -0)+\dfrac{216}{3}(1-1)]$

$V = [648(2 \pi )+\dfrac{216}{3}(0)]$

$V = 648(2 \pi )$

$\mathbf{V = 1296 \pi }$