# Volume by double integration

I had a question while finding the volume of the solid bounded "above" by $$f(x,y)=xy$$ and bounded "below" by $$x^2 + y^2 =9$$.

I used double integral in polar coordinate to calculate , then $$V = \int_0^{2\pi}\int_0^3 \cos\varphi\sin\varphi\ r^3 \,dr \,d\varphi$$.

Then i got $$V = \frac{81}{4}\int_0^{2\pi} \sin\varphi\cos\varphi \,d\varphi$$. And this integral equals to $$0$$.

Does this imply that the volume is $$0$$ or the right way to calculate the volume is $$V = \frac{81}{4}\cdot 4 \int_0^{\pi/2} \sin\varphi\cos\varphi \,d\varphi$$?

• Hint: what symmetries does the volume have?
– J.G.
Apr 14 at 6:56
• Please edit your question to include the differentials on your last integral. Apr 14 at 6:59
• Actually your 2nd integral is also missing the differentials. Apr 14 at 7:10
• Take a look at this graph. geogebra.org/3d/qqmqbkfs There's no need to even split into 4 integrals. One integral is fine. Apr 14 at 7:17
• @nickalh yes, the problem i have that i think that volume cannot be $0$. Apr 14 at 7:43

First look that the intersection of $$z=xy$$ with $$x^{2}+y^{2}=9$$ gives two regions above the $$z$$ axis and two regions below the $$z$$ axis. This corresponds to $$x,y$$ being in the (1st and 3rd quadrants) and $$x,y$$ being in the (2nd and 4th quadrants).

So by symmetry, you can just consider the 1st quadrant and multiply it by $$4$$.

In the first quadrant, you are looking at the volume bounded by $$z=r\cos(\theta)\sin(\theta)$$ for $$0\leq r\leq 3$$ and $$0\leq \theta\leq \frac{\pi}{2}$$

This gives you that $$\frac{V}{4}=\int_{0}^{3}\int_{0}^{\frac{\pi}{2}}r^{2}\cos(\theta)\sin(\theta)\cdot r\,dr\,d\theta=\frac{81}{8}$$

This gives you that $$V=\dfrac{81}{2}$$

The reason you are getting $$0$$ is because for $$\theta\in[\frac{\pi}{2},\pi]$$ and $$\theta\in[\frac{3\pi}{2},2\pi]$$, you get intersections in the negative $$z$$ axis which is giving you negative volume. So when you are adding all $$4$$, the integral will evaluate to $$0$$.

Or what you can do is integrate the modulus of the function $$xy$$.

Then you'll get $$V=\int_{0}^{3}\int_{0}^{2\pi}r^{3}|\cos(\theta)\sin(\theta)|\,dr\,d\theta$$

and this will give you $$V=\dfrac{81}{2}$$

• @user672587 Suppose you are integrating $\int_{-1}^{0} x\,dx$ . Then do you get a postivie value? A similar thing is also happening here. The reason simply being that $xy$ is negative in those quadrants. Thus, integrating a negative function over any region will give you a negative volume. That is why, if you are asked to find the area bounded by the curve $y=x$ and the $x$-axis for $-1\leq x\leq 1$, you either split it into two and write it as $2\cdot\int_{0}^{1}x\,dx$ or you write it as $\int_{-1}^{1}\,|x|\,dx$ because area cannot be a negative quantity. But still $\int_{-1}^{1}x\,dx=0$ Apr 14 at 9:01
• Oh yes, that's right. It's just that the way I expressed what I thought in the above comment was bad ((( Apr 14 at 9:18

To calculate a volume, we actually need absolute values on the integrand. $$V = \int_0^{2\pi}\int_0^3 |\cos\varphi\sin\varphi\ r^3 |\,dr \,d\varphi$$ becomes $$V = \frac{81}{4} \int_0^{2\pi} |\cos\varphi\sin\varphi\ | \ d\varphi$$ This is a pretty standard Calculus I integral. $$V = \frac{81}{4} \int_0^{2\pi} \frac12 \ | \sin2\varphi\ | \,d\varphi$$ Technically, we should split this into 8 integrals, but because each subinterval, $$[0,\frac\pi 2), [\frac\pi 2, \pi), [\pi, \frac {3\pi}{2}), [\frac{3\pi}{2}, 2\pi]$$ has the same result we merely need integrate one interval and multiply by 4. Skipping some steps, we arrive at $$V = \frac{81}{4}\ \frac12 \ \frac12 \ 4(1 -\ -1)= \frac{81}{4} \cdot 2=\frac{81}{2}$$ This is a reasonable result confirmed by intuition when we look at the graph. I've linked it which will allow you to rotate it and see it from various perspectives.

• The reason you are getting $0$ "volume" is that the integral is evaluating the volume of the regions in the 2nd and 4th quadrant to be negative. What you need to do is focus on the 1st and 3rd octants and multiply what you get by $2$ or you can focus on the 1st quadrant only and multiply what you get by $4$. The problem is occuring because you are taking $\theta\in[0,2\pi]$. Obviously, the volume is not zero. Apr 14 at 8:04
• So either you evaluate the integral of the absolute value of the integrand or you focus on the positive and negative parts separately and then add the absolute values of the result that you get in each region. Apr 14 at 8:15
• Thank you you two so much. I got it now. Apr 14 at 8:20
• Oops, I assumed "net volume" instead of simply volume. I've almost got a correction posted. Apr 14 at 8:24
• @nickalh You still need to edit it because you still have a typo of $1-\,-1=0$. Apr 14 at 8:56