1. ## Could someone explain how to answer these 2 multiple integration questions please

Hi, could someone explain step by step how to answer these questions please:

Calculate the volume:
1) z^2 + x^2 =1 and x+y+z=3, y=0

2) (x^2)/9 +(y^2)/4 =1, z=xy in the 1st quadrant

Thanks

2. Originally Posted by Summit
Hi, could someone explain step by step how to answer these questions please:

Calculate the volume:
1) z^2 + x^2 =1 and x+y+z=3, y=0

Thanks
Always draw a graph it helps you see the volume that you are trying to find.

The equation $z^2 + x^2 =1$ is a circluar cylindar of radius 1 centered on the y axis. It's bottom is the xz plane(y=0) and it is cut by the plane $x+y+z=3$.

To find the volume I want to use cylindrical coordinates, but since the cylindar is tipped over lets use this change of coordinates.

$x=r\cos(\theta) \\\ z=r\sin(\theta) \\\ y=y$

Now we get the integral

$\int_{0}^{2 \pi} \int_{0}^{3} \int_{0}^{3-r\cos(\theta)-3\sin(\theta)}rdydrd\theta$

You could have also set up a double integral using the same idea where y is your height function.

I hope this helps.

Good luck.

3. Originally Posted by Summit
Hi, could someone explain step by step how to answer these questions please:

2) (x^2)/9 +(y^2)/4 =1, z=xy in the 1st quadrant

Thanks
Once again draw a graph.

The equation $\frac{x^2}{9}+\frac{y^2}{4}=1$

Is an equation of an elliptic cylindar centered on the z axis. We are trying to find the volume under the surface z=xy inside the cylindar and in quadrant one.

the projection of the cylindar into the xy plane is the ellipse

$\frac{x^2}{9}+\frac{y^2}{4}=1$

We can islolate y in this expression to get

$y=\frac{2}{3}\sqrt{9-x^2}$

$\int_{0}^{x} \int_{0}^{\frac{2}{3}\sqrt{9-x^2}}xy \\\ dy \\\ dx$