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
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
Always draw a graph it helps you see the volume that you are trying to find.
The equation $\displaystyle z^2 + x^2 =1$ 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 $\displaystyle x+y+z=3$.
To find the volume I want to use cylindrical coordinates, but since the cylindar is tipped over lets use this change of coordinates.
$\displaystyle x=r\cos(\theta) \\\ z=r\sin(\theta) \\\ y=y$
Now we get the integral
$\displaystyle \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.
Once again draw a graph.
The equation $\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$
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
$\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$
We can islolate y in this expression to get
$\displaystyle y=\frac{2}{3}\sqrt{9-x^2}$
In the first quadrant
Now the inegral is
$\displaystyle \int_{0}^{x} \int_{0}^{\frac{2}{3}\sqrt{9-x^2}}xy \\\ dy \\\ dx$
Good luck.