# Please refer to image to populate this question textCalculate the volume of the region bounded by...

## Question:

Calculate the volume of the region bounded by

{eq}z=0, x^2 + y^2 =1 {/eq} and {eq}z = ax + by + 1 {/eq}.

It is assumed that {eq}ax+by+1 \geq 0 {/eq} when {eq}x^2 + y^2 \leq 1 {/eq}.

## Calculating the Volume:

The objective is to calculate the volume of the region.

The general form of volume is {eq}\displaystyle V = \iiint_{E} dV {/eq}

By using cylindrical coordinates we have to find the limits for integration and get a solution.

The given regions are:

{eq}\displaystyle z = 0 \\ \displaystyle x^2 + y^2 = 1 \\ \displaystyle z = ax + by + 1 {/eq}

Given that, assume

{eq}\displaystyle ax + by + 1 \geq 0 \\ \displaystyle x^2 + y^2 \leq 1 {/eq}

Therefore,

{eq}\displaystyle ax + by + 1 = 0 \\ \displaystyle x^2 + y^2 = 1 {/eq}

By using cylindrical coordinates:

{eq}\displaystyle x = r\cos \theta, \displaystyle y = r\sin \theta \\ \displaystyle x^{2} + y^{2} = \left ( r\cos \theta \right ) ^{2} + \left ( r\sin \theta \right ) ^{2} \\ \displaystyle x^{2} + y^{2} = r^{2} \left ( \cos^{2} \theta + \sin^{2} \theta \right ) \\ \displaystyle x^{2} + y^{2} = r^{2} {/eq}

Let us, find the limits:

{eq}\displaystyle x^2 + y^2 = 1 \\ \displaystyle r^2 = 1 \\ \displaystyle r = 1 \\ \displaystyle z = ax + by + 1 \\ \displaystyle z = a r \cos \theta + b r \sin \theta + 1 {/eq}

The limits are:

{eq}\displaystyle 0 \leq z \leq a r \cos \theta + b r \sin \theta + 1 \\ \displaystyle 0 \leq r \leq 1 \\ \displaystyle 0 \leq \theta \leq 2\pi {/eq}

Now, we are going to calculate the volume:

{eq}\begin{align*} \displaystyle V &= \iiint_{E} dV \\ \displaystyle &= \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{a r \cos \theta + b r \sin \theta +1} r dz dr d\theta \\ \displaystyle &= \int_{0}^{2\pi} \int_{0}^{1} \left [ rz\right ]_{0}^{a r \cos \theta + b r \sin \theta +1} dr d\theta \\ \displaystyle &= \int_{0}^{2\pi} \int_{0}^{1} \left [a r \cos \theta + b r \sin \theta +1 - 0 \right ] dr d\theta \\ \displaystyle &= \int_{0}^{2\pi} \int_{0}^{1} \left [ a r \cos \theta + b r \sin \theta +1 \right ] dr d\theta \\ \displaystyle &= \int_{0}^{2\pi} \left [ \frac{1}{2}r^2\left(1+ar\cos \theta +br\sin \theta \right)-\frac{r^3\left(a\cos \theta +b\sin \theta \right)}{6} \right ]_{0}^{1} d\theta \\ \displaystyle &= \int_{0}^{2\pi} \left [ \frac{1}{2}\cdot 1^2\left(1+a\cdot 1\cdot \cos \theta +b\cdot 1\cdot \sin \theta \right)-\frac{1^3\left(a\cos \theta +b\sin \theta \right)}{6} - 0\right ] d\theta \\ \displaystyle &= \int_{0}^{2\pi} \left [ \frac{1}{2}\left(1+a\cos \theta +b\sin \theta \right)-\frac{1}{6}\left(a\cos \theta +b\sin \theta \right) \right ] d\theta \\ \displaystyle &= \int_{0}^{2\pi} \left [ \frac{2a\cos \theta +2b\sin \theta +3}{6} \right ] d\theta \\ \displaystyle &= \left [ \frac{1}{6}\left(2a\sin \theta -2b\cos \theta +3\theta \right) \right ]_{0}^{2\pi} \\ \displaystyle &= \left [ \frac{1}{6}\left(2a\sin \left(2\pi \right)-2b\cos \left(2\pi \right)+3\cdot 2\pi \right) - \left ( \frac{1}{6}\left(2a\sin \left(0\right)-2b\cos \left(0\right)+3\cdot 0\right) \right ) \right ] \\ \displaystyle &= \left [ \frac{2\sin \left(2\pi \right)a-2\cos \left(2\pi \right)b+6\pi }{6} - \left ( \frac{\left(2a\sin \left(0\right)-2b\cos \left(0\right)+3\cdot 0\right)}{6} \right ) \right ] \\ \displaystyle &= \left [ \frac{2\left(3\pi -b\right)}{6} - \left ( -\frac{2b}{6} \right ) \right ] \\ \displaystyle &= \left [ \frac{3\pi -b}{3}-\left(-\frac{b}{3}\right) \right ] \\ \displaystyle &= \left [ \pi - \frac{b}{3} + \frac{b}{3} \right ] \\ \displaystyle \therefore V &= \pi \end{align*} {/eq}

Therefore, the volume is {eq}\displaystyle \pi \ cubic \ units {/eq}.