Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the volume of the solid bounded by the coordinate planes and the plane $8x + 7y + z = 56$.

The solution states: $$ V= \int_0^9 \int_0^{8-\frac{8}{9}x} 72-8x-9y \; dy\; dx$$


But I am unsure as to why the integration limit is what it is especially the limit of integration on $x$. Can anyone clarify?

share|improve this question
add comment

2 Answers 2

I can certainly see why you're confused! It looks like you've looked at the wrong problem (or perhaps the right problem, but the wrong problem's answer). That double integral is what would come from a similar situation, but with your plane having equation $8x+9y+z=72$ instead of $8x+7y+z=56$.

To get the integrand, we can solve $8x+7y+z=56$ for $z$ (we needn't do anything else, since the volume is bounded by the $xy$-plane, rather than some more complicated surface). For the inner integral, we start with the equation $8x+7y+z=56$, then set $z=0$ and solve the resulting linear equation for $y$ to give us the upper limit (the lower, $y=0$, comes from the fact that the volume is bounded by the $xz$-plane). For the outer integral, we start with the linear equation we got in the previous step, then set $y=0$ and solve the resulting equation for $x$ to get our upper limit (again, the lower limit comes from the coordinate plane bounding).

share|improve this answer
add comment

Since the region is bounded by the coordinate axes, begin by solving the expression for $z$ (although you could integrate over any of the three variables first as they play essentially symmetric roles). This gives $$ z = 56 - 8x - 7y $$ Thus, our limits for the integration over $z$ are given by $$ 0 \leq z \leq 56 - 8x - 7y $$ Now we consider the region in the $xy$-plane by setting $z = 0$. $$ 8x + 7y = 56 $$ Solving this for $y$, we get $$ y = 8 - \frac{8}{7}x $$ so the bounds of integration for $y$ are given by $$ 0 \leq y \leq 8 - \frac{8}{7}x $$ We now consider where this curve crosses the $x$-axis by setting $y = 0$. $$ \frac{8}{7}x = 8 $$ $$ x = 7 $$ Thus, the bounds of integration for $x$ are given by $$ 0 \leq x \leq 7 $$ Putting this all together gives us $$ \int_{0}^{7} \int_0^{8 - \frac{8}{7}x} \int_0^{56 - 8x - 7y} dz \, dy \, dx $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.