I understand that this can be done with triple integrals, but my class has yet to be taught those and we will be assessed on our ability to perform a question similar to this one with the principles of double integrals, so, how would the volume of this solid be found using double integrals?
I was thinking of using Green's Theorem, but I could not find a way to make it work for this problem.
Maybe a line integral would work, but I do not know how to set one up in this scenario.
Best Answer
Maybe it's interesting to write the triple integral and see how it transforms in a double one if we integrate first wrt $z$.
As the line $x+y=4$ cuts the $y$ axis at $x=4$, $x$ ranges from $0$ to $4$. Now, for a given value of $x$, $y$ ranges from $0$ to $4-x$. For each $x$ and $y$, $z$ varies from $0$ to $\sqrt{4-x}$. The volume element is $\mathbb dz\,\mathbb dy\,\mathbb dx$
$$V=\int_0^4\int_0^{4-x}\int_0^{\sqrt{4-x}}\mathbb dz\,\mathbb dy\,\mathbb dx=$$
$$=\int_0^4\int_0^{4-x}\sqrt{4-x}\,\mathbb dy\,\mathbb dx$$
We can arrive directly to the last expression considering that the integration of the function $z=f(x,y)=\sqrt{4-x}$ over some region implies vertical surfaces, the ones given by the wording and this integral is the volume enclosed by those vertical surfaces and the surface of the function.