[size=150][b]Double integrals on Type I regions[/b][/size][br][br]Suppose [math]f(x,y)[/math] is defined on [math]D[/math], a closed and bounded region in [math]\mathbb{R}^2[/math]. How can we define the double integral of [math]f[/math] over [math]D[/math]? [br][br]First, we consider [math]D[/math] to be the type of regions such that the upper boundary is the graph of [math]y=h(x)[/math] and the lower boundary is the graph of [math]y=g(x)[/math] for [math]a\leq x \leq b[/math]. [math]D[/math] is called a [b]Type I region[/b]. Then [math]\int_{g(x)}^{h(x)} f(x,y) \ dy[/math] can be regarded as a "slice" of the solid under the surface i.e. the graph of [math]z=f(x,y)[/math] over [math]D[/math] for a fixed [math]x[/math]. Then we can compute the volume of the solid under the surface by integrating these slices for all [math]x[/math] from [math]a[/math] to [math]b[/math]. Therefore, the double integral is equal to the following iterated integral:[br][br][math]\iint_D f(x,y) \ dA=\int_a^b \left(\int_{g(x)}^{h(x)} f(x,y) \ dy\right) \ dx[/math][br][br][u]Example[/u]: Suppose [math]f(x,y)=y^2x+1[/math] on the domain [math]D[/math] bounded by the graphs of [math]y=-x[/math] and [math]y=x^2[/math] and the vertical line [math]x=1[/math]. Compute [math]\iint_D f(x,y) \ dA[/math].[br][br][u]Answer[/u]: [br][br]First of all, the graphs of [math]y=-x[/math] and [math]y=x^2[/math] intersect at [math](0,0)[/math], which means that the range of [math]x[/math] is from [math]0[/math] to [math]1[/math]. Moreover, [math]x^2\geq -x[/math] on [math][0,1][/math]. Therefore, we have [math]g(x)=-x[/math] and [math]h(x)=x^2[/math]. The double integral can be computed as follows:[br][br][math]\iint_R f(x,y) \ dA=\int_0^1\left(\int_{g(x)}^{h(x)}f(x,y) \ dy\right) \ dx = \int_0^1\left(\int_{-x}^{x^2}(y^2x+1) \ dy\right) \ dx[/math][br][math]=\int_0^1\left[\frac{y^3x}3+y\right]_{-x}^{x^2} \ dx = \int_0^1\left(\frac{x^7}3+\frac{x^4}3+x^2+x\right) \ dx[/math][br][math]=\left[\frac{x^8}{24}+\frac{x^5}{15}+\frac{x^3}3+\frac{x^2}2\right]_0^1=\frac{113}{120}[/math][br][br]In the applet below, the left panel shows the region [math]D[/math] with the graph of [math]y=h(x)[/math] as upper boundary and the graph of [math]y=g(x)[/math] as lower boundary for [math]a\leq x \leq b[/math]. The right panel shows the solid under the graph of [math]z=f(x,y)[/math] over [math]D[/math]. Also, the "slice" of the solid with fixed [math]x[/math] is also shown. You can drag the slider to change the value of [math]x[/math].[br][br]

[b][size=150]Double integrals on Type II regions[/size][/b][br][br]Now we consider [math]E[/math] to be another type of regions such that the left boundary is the graph of [math]x=p(y)[/math] and the right boundary is the graph of [math]x=q(y)[/math] for [math]c\leq y \leq d[/math]. [math]E[/math] is called a [b]Type II region[/b]. Using the similar idea, we can compute the volume of the solid under the graph of [math]z=f(x,y)[/math] over [math]E[/math] by integrating the slices over all [math]y[/math] from [math]c[/math] to [math]d[/math] and express the double integral as the following iterated integral:[br][br][math]\iint_E f(x,y) \ dA = \int_c^d \left(\int_{p(y)}^{q(y)} f(x,y) \ dx\right) \ dy[/math][br][br][br][u]Example[/u]: Let [math]f(x,y)=x\sin y[/math]. Compute [math]\iint_E f(x,y) \ dA[/math], where [math]E[/math] is the region bounded by [math]x=0[/math] and [math]x=\cos(y)[/math] for [math]0\leq y \leq \frac{\pi}3[/math]. [br][br][u]Answer[/u]:[br][br]Since [math]\cos(y)\geq 0[/math] on [math]\left[0,\frac{\pi}3\right][/math], we have [math]p(y)=0[/math] and [math]q(y)=\cos(y)[/math]. The double integral can be computed as follows:[br][br][math]\iint_E f(x,y) \ dA=\int_0^{\frac{\pi}3}\left(\int_{p(y)}^{q(y)} f(x,y) \ dx\right) \ dy=\int_0^{\frac{\pi}3}\left(\int_0^{\cos(y)}x\sin y \ dx\right) \ dy[/math][br][math]=\int_0^{\frac{\pi}3}\left[\frac{x^2}2\sin(y)\right]_0^{\cos(y)} \ dy=\int_0^{\frac{\pi}3}\frac{\cos^2(y)\sin(y)}2 \ dy[/math][br][math]=\left[-\frac{\cos^3(y)}6\right]_0^{\frac{\pi}3}=-\frac 1{48}+\frac 16=\frac 7{48}[/math][br][br]

[b][size=150]Solid between two surfaces[/size][/b][br][br]Given two functions [math]f(x,y)[/math] and [math]g(x,y)[/math] such that [math]f(x,y)\geq g(x,y)[/math] on a region [math]S[/math]. We can compute the volume [math]V[/math] between the graphs of [math]z=f(x,y)[/math] and [math]z=g(x,y)[/math] over the region [math]S[/math] by the following double integral:[br][br][math]V=\iint_S (f(x,y)-g(x,y)) \ dA[/math][br][br][br][u]Example[/u]: Find the volume of the solid bounded by [math]z=x^2+y^2[/math] from below and [math]z=8-(x^2+y^2)[/math] from above.[br][br][u]Answer[/u]:[br][br]First, we need to find the intersection curve of the two given surfaces. By solving the two given equations, we get[br][br][math]x^2+y^2 =8-(x^2+y^2) \implies x^2+y^2=4[/math][br][br]In other words, the intersection curve is a circle as shown in the applet below. Let [math]C=\left\{(x,y) \ | \ x^2+y^2\leq 4\right\}[/math]. The volume [math]V[/math] bounded by the two surfaces is as follows:[br][br][math]V=\iint_C (8-(x^2+y^2)-(x^2+y^2)) \ dA=\iint_C (8-2x^2-2y^2) \ dA[/math][br][br]Since the solid is rotationally symmetric, we can consider the double integral over the the quarter-disk [math]Q[/math] in the first quadrant, bounded by [math]y=0[/math] and [math]y=\sqrt{4-x^2}[/math] for [math]0\leq x\leq 2[/math]. Therefore, we have[br][br][math]V=4\iint_Q (8-2x^2-2y^2) \ dA[/math][br][math]=4\int_0^2\left(\int_0^{\sqrt{4-x^2}}(8-2x^2-2y^2) \ dy\right) \ dx[/math][br][math]=8\int_0^2\left(\int_0^{\sqrt{4-x^2}}(4-x^2-y^2) \ dy\right) \ dx[/math][br][math]=8\int_0^2\left[(4-x^2)y-\frac{y^3}3\right]_0^{\sqrt{4-x^2}} \ dx[/math][br][math]=\frac{16}3\int_0^2(4-x^2)^{\frac 32} \ dx[/math][br][br]Use the substitution [math]x=2\sin\theta[/math], we get[br][br][math]V=\frac{256}3\int_0^{\frac{\pi}2} \cos^4\theta \ d\theta=\frac{256}3\int_0^{\frac{\pi}2} \left(\frac{1+\cos(2\theta)}2\right)^2 \ d\theta=\frac{256}3\int_0^{\frac{\pi}2} \frac{1+2\cos(2\theta)+\cos^2(2\theta)}4 \ d\theta[/math][br][math]=\frac{256}3\int_0^{\frac{\pi}2} \frac{3+4\cos(2\theta)+\cos(4\theta)}8 \ d\theta=\frac{32}3\left[3\theta+2\sin(2\theta)+\frac{\sin(4\theta)}4\right]_0^{\frac{\pi}2}=16\pi[/math][br]