I recently learned Multiple integrals and these problems are from a past-paper. enter image description here

Please someone point me to the right direction where should I start to solve these problems? I have looked over many sources but I can't seem to find these kind of problems with solutions.

In the first problem I drew an sketch of it and found out that it has a triangle shape, but there is another triangle shape inside the big triangle. I'm confused what does it means by "area enclosed by the curves", so I'm not sure how should I find the limits to the integral. (I'm not a native english speaker)

In the second problem after I drew a sketch I and realized that I have to find the volume but the half of cylinder get cuts by the 'y=0'.

I know I have to integrate z=4-y by dxdy I also have found that the limits for dx is -3 and 3 since it's circles equation, so I don't know how to proceed further.

What are the things I should consider first when I get problems like this? Please point me to the right direction anyone.

Must I always integrate z when I get to find volume?


1 Answer 1


Hint for a). The domain of integration is a trapezoid of vertices $(0,0)$, $(1/2,4)$, $(2,4)$ and $(6,0)$. Then the double integral can be written as an iterated integral $$\int_{y=0}^{4}\left(\int_{x=?}^{?}(x^2+y^2)dx\right)dy$$ Can you take it from here?

Hint for b). Yes here the solid is a cylinder $x^2+y^2=9$ cut by the planes $y=0$, $z=0$ and $z=4-y$. Note that the base of this solid is given by the disc $D:=\{(x,y): x^2+y^2\leq 3^2\}$ which is cut into two parts by the plane $y=0$: $$D^+=D\cap \{y\geq 0\}\quad\mbox{and}\quad D^-=D\cap \{y\leq 0\}.$$ So, here we have two bounded parts whose volumes are respectively $$V_+=\int_{(x,y)\in D^+}\left(\int_{z=0}^{4-y}dz\right)dxdy,\quad V_-=\int_{(x,y)\in D^-}\left(\int_{z=0}^{4-y}dz\right)dxdy.$$

  • $\begingroup$ thank you very much for the answer. ah, I get it now, I was wrong about the 'enclosed', so it's shape is trapezoid. So the x is from 0 to 6. Thank you again. $\endgroup$ Sep 13, 2017 at 13:05
  • $\begingroup$ @user2511652 No. the missing limits in the integral are functions $x=f(y)$. Which one? $\endgroup$
    – Robert Z
    Sep 13, 2017 at 13:07
  • $\begingroup$ thank you again sir. for the answer b. can you please explain me more about the limits of y $\endgroup$ Sep 13, 2017 at 13:07
  • $\begingroup$ ah, I meant the first problem. the limits of x $\endgroup$ Sep 13, 2017 at 13:08
  • $\begingroup$ @user2511652 Take the trapezium and draw horizontal lines $y=k$ with $k\in [0,4]$. Each line intersects the sides of the trapezium. The left side gives you the lower limit. The right side gives you the upper limit. $\endgroup$
    – Robert Z
    Sep 13, 2017 at 13:10

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