# Find the Volume (Double Integral)

## Main Question or Discussion Point

I'm having trouble trying to setup this double integral. The question asks to find the volume of a solid enclosed by the parabolic cylinder $$y = x^{2}$$ and the planes $$z = 3y$$, $$z = 2+y$$

I'm not even sure where to start. I have drawn the figure and understand that you have to integrate the two functions $$z = 3y$$ and $$z = 2+y$$ and subtract the volumes. However I'm stuck trying to setup the boundaries. Thanks.

To find a volume I suppose you have to set up a triple integral?

You may, for example, let:
$$3y < z < 2 + y, \quad -\sqrt{y} < x < \sqrt{y} \quad, 0 < y < 1$$

.. I think, not sure, though.

I'm having trouble trying to setup this double integral. The question asks to find the volume of a solid enclosed by the parabolic cylinder $$y = x^{2}$$ and the planes $$z = 3y$$, $$z = 2+y$$

I'm not even sure where to start. I have drawn the figure and understand that you have to integrate the two functions $$z = 3y$$ and $$z = 2+y$$ and subtract the volumes. However I'm stuck trying to setup the boundaries. Thanks.

If the problem is correct as you stated it the boundaries would be $0\leq y\leq \Gamma$ and $-\sqrt{y}\leq x\leq\sqrt{y}$, where $\Gamma$ ist the y-value for which the two planes intersect.

So

$$\int_0^\Gamma{dy\int_{-\sqrt{y}}^{\sqrt{y}}dx(2y-2)}$$

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