Section 4.3 Double Integrals Polar
4.3 Double Integrals Polar
In previous section, we learn the double integral over general regions using rectangular coordinates [latex](x,y)[/latex]. This is in general not very practical if we have a region that is circular, for example a circle. It is naturally that we use the polar coordinate for the integration as we did in calculus II. Recall that any point [latex](x,y)[/latex] in the plane, we can convert it into [latex](r,\theta)[/latex] where [latex]x^{2}+y^{2}=r^{2}[/latex], [latex]x=r\text{cos}(\theta)[/latex] and [latex]y=\text{sin}(\theta)[/latex]. Given
\[R=\{(r,\theta)|a\leq r\leq b,\alpha\leq\theta\leq\beta\},\]
a polar region, one can find the volume between a continuous function [latex]f(x,y)\geq0[/latex] and [latex]xy[/latex]-plane with base [latex]R[/latex] using the polar coordinate system.
Definition: Polar Coordinates in a Double Integral
The double integral of the function [latex]f(r,\theta)[/latex] over the polar rectangular region [latex]R[/latex] in the [latex]r\theta[/latex] -plane is defined as
\[\int\int_{R}f(r,\theta)dA=\lim_{m,n\rightarrow\infty}\sum_{i=1}^{m}\sum_{j=1}^{n}f(r_{ij}^{*},\theta_{ij}^{*})\triangle A\]
\[ =\lim_{m,n\rightarrow\infty}\sum_{i=1}^{m}\sum_{j=1}^{n}f(r_{ij}^{*},\theta_{ij}^{*})r_{ij}^{*}\triangle r\triangle\theta\]
\[ =\int_{\theta_{1}}^{\theta_{2}}\int_{r_{1}}^{r_{2}}f(r,\theta)rdrd\theta.\]
When the function [latex]f[/latex] is given in terms of [latex]x[/latex] and [latex]y[/latex], using [latex]x=r\text{cos}(\theta)[/latex] and [latex]y=\text{sin}(\theta)[/latex], and [latex]dA=rdrd\theta[/latex] changes it to
\[ \int\int_{R}f(x,y)dA=\int_{\theta_{1}}^{\theta_{2}}\int_{r_{1}}^{r_{2}}f(r\text{cos}(\theta),r\text{sin}(\theta))rdrd\theta. \]
Example 1: Compute the double integral [latex]\iint_{R}f(x,y)dA[/latex] where [latex]f(x,y)=y[/latex] and [latex]R=\{(r,\theta)|1\leq r\leq2,0\leq\theta\leq\frac{\pi}{2}\}[/latex]
Exercise 1: Compute the double integral [latex]\iint_{R}f(x,y)dA[/latex] where [latex]f(x,y)=x[/latex] and [latex]R=\{(r,\theta)|2\leq r\leq3,\frac{\pi}{3}\leq\theta\leq\frac{3\pi}{4}\}[/latex]
Example 2: Compute the double integral [latex]\iint_{R}1-x^{2}-y^{2}dA[/latex] where [latex]R[/latex] is the unit circle on the [latex]xy[/latex]-plane.
Exercise 2: Compute the double integral [latex]\iint_{R}x^{2}+y^{2}dA[/latex] where [latex]R[/latex] is a circle with radius [latex]2[/latex] on the [latex]xy[/latex]-plane.
Theorem: Double Integrals over General Polar Regions
If [latex]f(r,\theta)[/latex] is continuous on a general polar region
\[R=\{(r,\theta)|h_{1}(\theta)\leq r\leq h_{2}(\theta),\alpha\leq\theta\leq\beta\},\]
then
\[\int\int_{R}f(r,\theta)dA=\int_{\alpha}^{\beta}\int_{h_{1}(\theta)}^{h_{2}(\theta)}f(r,\theta)rdrd\theta.\]
Example 3: Compute the double integral [latex]\iint_{R}1dA[/latex] where [latex]R=\{(r,\theta)|0\leq r\leq3\text{cos}(\theta),\frac{\pi}{3}\leq\theta\leq\frac{2\pi}{3}\}[/latex].
Exercise 3: Compute the double integral [latex]\iint_{R}2dA[/latex] where [latex]R=\{(r,\theta)|0\leq r\leq\text{cos}(\theta),\frac{\pi}{4}\leq\theta\leq\frac{3\pi}{4}\}[/latex].
Example 4: Compute the double integral [latex]\iint_{R}x+2ydA[/latex] where [latex]R[/latex] is in the second quadrant enclosed by [latex]x^{2}+y^{2}=9[/latex], [latex]x=0[/latex] and [latex]y=-x[/latex].
Exercise 4: Compute the double integral [latex]\iint_{R}2x-ydA[/latex] where [latex]R[/latex] is in the first quadrant enclosed by [latex]x^{2}+y^{2}=4[/latex], [latex]y=0[/latex] and [latex]y=x[/latex].
Example 5: Compute the double integral [latex]\iint_{R}e^{-x^{2}-y^{2}}dA[/latex] where [latex]R[/latex] is in the third quadrant between the circles with center on the origin and radii [latex]2[/latex] and [latex]3[/latex].
Exercise 5: Compute the double integral [latex]\iint_{R}e^{x^{2}+y^{2}}dA[/latex] where [latex]R[/latex] is in the second quadrant between the circles with center on the origin and radii [latex]1[/latex] and [latex]2[/latex].
Example 6: Find the volume under the paraboloid [latex]z=x^{2}+y^{2}[/latex] and above [latex]9\leq x^{2}+y^{2}\leq25[/latex].
Exercises 6: Find the volume under the paraboloid [latex]z=-x^{2}-y^{2}+4[/latex] and above [latex]1\leq x^{2}+y^{2}\leq4[/latex].
Example 7: Evaluate
\[\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}e^{-x^{2}-y^{2}}dydx\]
where [latex]0\leq y\leq\sqrt{4-x^{2}}[/latex] and [latex]0\leq x\leq2[/latex].
Example 8: Find the volume of the solid above the cone [latex]z=\sqrt{x^{2}+y^{2}}[/latex] and below the sphere [latex]x^{2}+y^{2}+z^{2}=1[/latex].
Group work:
1. Evaluate
\[ \int_{0}^{3}\int_{0}^{\sqrt{9-y^{2}}}e^{x^{2}+y^{2}}dxdy \]
where [latex]0\leq x\leq\sqrt{9-y^{2}}[/latex] and [latex]0\leq x\leq1[/latex].
2. Find the volume of the solid bounded by paraboloid [latex]z=6-x^{2}-y^{2}[/latex] and [latex]z=x^{2}+y^{2}[/latex].
3. Find the volume of the solid inside [latex]x^{2}+y^{2}+z^{2}=25[/latex] and outside the cylinder [latex]x^{2}+y^{2}=9[/latex].
4. Find the volume of the solid bounded by paraboloid [latex]z=1+x^{2}+y^{2}[/latex] and [latex]z=5[/latex].