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The Plane: $x+y=4$
The Cylinder: $y^2+9z^2=16$

I have gone this far but I'm not sure it's true $$V=\int\limits_0^4\int\limits_0^{4-x}\int\limits_0^{\sqrt{16-y^2}/9} dz\ dy\ dx$$ PS: Answer can contain $\pi$ if needed.

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2 Answers

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\begin{align} V&=\int_0^4\int_0^{4-x}\int_0^{\sqrt{16-y^2}/3} dzdydx \\ &= \int_0^4\int_0^{4-y}\int_0^{\sqrt{16-y^2}/3} dzdxdy \\ &= \int_0^4 (4-y) \, \frac{\sqrt{16-y^2}}3 dy \\ &= \frac13 \left[ 4\int_0^4 \sqrt{16-y^2} dy - \int_0^4 y\sqrt{16-y^2} dy \right] \\ &= \frac13 \left[ 4 \, \frac{\pi(4^2)}{4} + \frac12 \int_0^4 \sqrt{16-y^2} d(16-y^2) \right] \\ &= \frac13 \left[ 4 \, \frac{\pi(4^2)}{4} - \frac12 \,\frac23 \, 16^{3/2} \right] \\ &= \frac13 \left( 16\pi - \frac{64}{3} \right) \\ &= \frac{16(3\pi-4)}{9} \end{align}

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Comparison of my answer with another answer

Since pictures are used for illustration, let me also use them.

1. Another answer: fix $x$ first. This gives rise to a more complicated integrand $$V = \frac13\int_0^4 \int_0^{4-x} \sqrt{16-y^2} dydx$$ As the shaded area of the graph illustrates, one needs to sum the area of a triangle and the sector (involving inverse trigonometric function). That's why wolfram-alpha complains "standard computation time exceeded...".
   
2. My answer: fix $y$ first. This change of order of integration kills two inner integrals, leaving a much simpler integrand in $y$ only. This allows wolfram-alpha to give the exact solution.
   

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Yes it is (almost) correct, indeed we need to consider that the plane intersect the cylinder according to the following sketch, then the set up is as follow

$$V=\int_0^4 dx\int_0^{4-x}dy\int_0^{\sqrt{\frac{16-y^2}{9}}} dz$$

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