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Summary of double integrals in Cartesian coordinates

(Cartesian coordinates are just $(x,y)$ in 2D or $(x,y,z)$ in 3D)

If you have a rectangular region of integration $D=\{(x,y)\mid a\leq x\leq b,~ c\leq y\leq d\}$ Then: $$\iint_D f(x,y)dA=\int_a^b\int_c^df(x,y)dy\,dx=\int_c^d\int_a^bf(x,y)dx\,dy$$

For a Type I region of integration: $$D=\{(x,y)\mid a\leq x\leq b,~ g_1(x)\leq y\leq g_2(x)\}$$ $$\iint_D f(x,y)dA=\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)dydx$$

For a Type II region of integration: $$D=\{(x,y)\mid c\leq y\leq d,~ h_1(y)\leq x\leq h_2(y)\}$$ $$\iint_D f(x,y)dA=\int_c^d\int_{h_1(y)}^{h_2(y)}f(x,y)dxdy$$

Example 1

Question: calculate the volume of the 3D body between $z=f(x,y)=(2x+3)e^{2y}$ and the $xy$-plane, when the bounds of $x$ and $y$ are the rectangle $-1\leq x\leq1$ and $0\leq y\leq2$.

Solution: since the region of integration is a rectangle, we could do both orders of integration, and they should give the same result. In this case, it is easier to immediately split up the integral into the product of two single integrals:
\begin{align}\text{Volume}&=\int_{-1}^{1}\red{\int_{0}^{2}(2x+3)e^{2y} dy}\,dx=\left(\int_{-1}^1(2x+3)dx\right)\left(\int_0^2 e^{2y}dy\right)\nonumber\\ &=\left[x^2+3x\right]_{x=-1}^{x=1}\left[\frac{1}{2}e^{2y}\right]_{y=0}^{y=2}=6\left(\frac{1}{2}(e^4-1)\right)=\boxed{3(e^4-1)}\nonumber \end{align}

Example 2

Question: calculate the volume of the 3D body between the paraboloid $z=x^2+y^2$ and the $xy$-plane, above the region $D$ enclosed by the parabola $y=3x^2$ and the line $y=x+2$.
Solution: in this case it is useful to make a (qualitative) sketch. We know that it looks somewhat like this:

Now, we can solve the equation $3x^2=x+2$ to see that the two graphs intersect at the points $\left(-\frac{2}{3},\red{\frac{4}{3}}\right)$ and $\left(1,\blue3\right)$.

Now, the question is; do we use a Type I or Type II region to integrate over?
Well, we see that if we let $x$ run from $-\frac{2}{3}$ to $1$, then for all $x$, the bounds of $y$ are already nicely given in terms of $x$, namely: $3x^2\leq y\leq x+2$. So, the bounds of $y$ depend on $x$. Therefore, it would make sense to use Type I. (We will perform the Type I calculation, after first explaining why Type II is not convenient to choose)

Suppose that instead we were to use Type II integration. Then the boundaries would become $x=\pm\sqrt{y/3}$ for the parabola and $x=y-2$ for the line. But look: in the bottom of the region, $x$ runs from the left part of the parabola to the right part of the parabola, whereas in the top of the region, $x$ runs between the line and the right part of the parabola. If we really were to perform Type II integration, we would have to split up the region into two parts and we would get two integrals: $$\text{Volume}=\int_0^{\red{4/3}}\int_{-\sqrt{y/3}}^{\sqrt{y/3}}(x^2+y^2)dx\,dy+\int_{\red{4/3}}^{\blue3}\int_{y-2}^{\sqrt{y/3}}(x^2+y^2)dx\,dy$$ That does not look nice. So, that's why Type I integration is really better in this case.

Let us write down the region as a Type I region (where we let $x$ run from $-\frac{2}{3}$ to $1$ and $y$ between the parabola and the line) $$D=\{(x,y)\mid-\frac{2}{3}\leq x\leq1,~3x^2\leq y\leq x+2\}$$ Then the integral becomes: $$V=\iint_{D}(x^2+y^2)dA=\int_{-2/3}^{1}\int_{3x^2}^{x+2}(x^2+y^2)dydx$$ Which we can "just" solve: \begin{align} V&=\int_{-2/3}^{1}\int_{3x^2}^{x+2}(x^2+y^2)dydx=\int_{-2/3}^{1}\left[x^2y+\frac{y^3}{3}\right]_{y=3x^2}^{y=x+2}dx\nonumber\\ &=\int_{-2/3}^{1}\left[x^2(x+2)+\frac{1}{3}(x+2)^3-x^2\cdot3x^2-\frac{1}{3}(3x^2)^3\right]dx\nonumber\\ &=\int_{-2/3}^{1}\left[x^3+2x^2+\frac{1}{3}\left(x^3+6x^2+12x+8\right)-3x^4-9x^6\right]dx\nonumber\\ &=\int_{-2/3}^{1}\left(-9x^6-3x^4+\frac{4}{3}x^3+4x^2+4x+\frac{8}{3}\right)dx\nonumber\\ &=\left[-\frac{9}{7}x^7-\frac{3}{5}x^5+\frac{1}{3}x^4+\frac{4}{3}x^3+2x^2+\frac{8}{3}x\right]_{-2/3}^{1} =\boxed{ \frac{3125}{567}}\nonumber \end{align} Conclusion: the volume is $\boxed{ \frac{3125}{567}}$

Example 3

Question: evaluate $\displaystyle\int_0^1\int_{5x}^5e^{y^2}dydx$.

Solution: Well, it's basically impossible to find a "nice" antiderivative to $e^{y^2}$ with respect to $y$, a useful fact to remember. So we cannot evaluate the integral as it stands now. So what do we do?

Answer: we change from Type I to Type II!
As we can deduce from the integral, right now we're integrating over the Type I region $$D=\{(x,y)\mid 0\leq x\leq1,~5x\leq y\leq5\}$$ But this is a triangle, with vertices $(0,0)$, $(0,5)$ and $(1,5)$!
Realizing that the boundary line can be written as $y=5x$ or as $x=\frac{y}{5}$, we can rewrite the region as a Type II region: $$D=\left\{(x,y)\mid 0\leq y\leq5,~0\leq x\leq\frac{y}{5}\right\}$$ So we can rewrite the integral, just like we rewrote the region: \begin{align} \int_0^1&\int_{5x}^5e^{y^2}dydx=\iint_De^{y^2}dA=\int_0^5\int_0^{y/5}e^{y^2}dxdy = \int_0^5 \left[xe^{y^2}\right]_{x=0}^{x=y/5}dy\nonumber\\ &=\frac{1}{5}\int_0^5ye^{y^2}dy=\frac{1}{5}\left[\frac{1}{2}e^{y^2}\right]_0^5=\boxed{\frac{1}{10}(e^{25}-1)}\nonumber \end{align} Here we used the fact that although the antiderivative of $e^{y^2}$ with respect to $y$ is impossible to find, the antiderivative of $e^{y^2}$ with respect to $x$ is very easy (namely, $xe^{y^2}$). And then it suddenly became possible to solve the integral! :)