Find the volume of the solid below the plane $z = x + 3y$ and above the ellipse $25x^2 + 16y^2 = 400, \ x \geq 0, \ y \geq 0$.

Question

Here I am having trouble in finding the bounds of integration. Since we are given that $x \geq 0 \ \text{and}\\ y \geq 0$. we are concerned only about the first quadrant, right?

For finding the volume, I tried double integrating $f\left(x,y\right) = x+3y$.

I tried using polar coordinates but could not define the bounds for $\theta$.

Thank you.

Answer 1

You have good intuition.

As for the transformation, I suggest the generalized polar coordinates $x=4r\cos\theta$, $y=5r\sin\theta.$

The volume is equal to $\int_0^1 \int_0^{\pi/2} \int_0^{4r\cos\theta+15r\sin\theta}20r\; dz\; d\theta \;dr$.