Answers: Hour Test Two Math 2058: Fall 2011
Problem
Set up {\bf but do not evaluate} an integral equal to the volume enclosed by $z=x^2+3y^2$ and $z=8-x^2-y^2$.
When the surfaces intersect: 8-x^2-y^2 == (x^2+3y^2),
This projects to the interior of the region bounded by 8 = 2 x^2 + 4 y^2 (an ellipse) in the xy-plane
In[1]:=
Out[1]=
Integrate@Integrate@Integrate@1, 8z, x ^ 2 + 3 y ^ 2, 8 - x ^ 2 - y ^ 2<D,
8y, - Sqrt@2 - x ^ 2 ê 2D, Sqrt@2 - x ^ 2 ê 2D<D, 8x, - 2, 2<D
2 p
8
Note: Evaluation was not required
Problem
Evaluate the iterated integral: $$\int_0^1 \int_0^{\sqrt{1-x^2}} e^{x^2+y^2} \,dy\,dx$$
Answer: Change to polar coordinates
Integrate@Integrate@Exp@r ^ 2D r, 8r, 0, 1<D, 8t, 0, Pi ê 2<D
1
4
H- 1 + ‰L p
Problem
Use spherical coordinates to evaluate
\[\int_0^4 \int_0^{\sqrt{16-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{32-x^2-y^2}}
\sqrt{x^2+y^2+z^2}\,dz\,dy\,dx.\]
Answer: Integrand is r r^2 Sin[f]
Region: Projects to first quadrant interior of circle of radius 4 in xy-plane
"Ice cream" cone (z^2=x^2+y^2) solid inside sphere of radius Sqrt[32]
Integrate@Integrate@Integrate@r r ^ 2 Sin@fD, 8f, 0, Pi ê 4<D, 8r, 0, Sqrt@32D<D, 8t, 0, Pi ê 2<D
1
2
K256 - 128
2 Op
Problem
Consider the double iterated integral:
\int_0^1 \int_x^{\sqrt{x}} f(x,y) \,dy\,dx
Sketch (or describe in words) the region of integration $R$.
Evaluate the integral if $f(x,y) = 1$.
Setup {\bf but do not evaluate} an equivalent iterated integral with the order of integration reversed.
Region between x = 0 and x=1 above y=0 and below y=Sqrt[x]
2
11f58t2Ans.nb
In[2]:=
Plot@8x, Sqrt@xD<, 8x, 0, 1<D
1.0
0.8
0.6
Out[2]=
0.4
0.2
0.2
In[3]:=
0.4
0.6
0.8
1.0
Integrate@Integrate@1, 8y, x, Sqrt@xD<D, 8x, 0, 1<D
1
Out[3]=
6
Integrate[Integrate[F[x, y], {x, y^2, y}], {y, 0, 1}]
Now just to check:
In[4]:=
Integrate@Integrate@1, 8x, y ^ 2, y<D, 8y, 0, 1<D
1
Out[4]=
6
Problem
Find the double integral of $f(x,y)=x y$ over the triangular region with vertices $(1,0),(0,1),(0,-1)$.
Plot@81 - x, - 1 + x<, 8x, - 1, 2<D
2
1
-1.0
-0.5
0.5
1.0
1.5
2.0
-1
-2
Integrate@Integrate@x y, 8y, - 1 + x, 1 - x<D, 8x, 0, 1<D
0
Note: The inner integrand is an odd function in y and the limits are of the form {- a, a} hence the inner integral is zero.
Problem
11f58t2Ans.nb
Problem
Find the volume of the solid under the graph of $f(x,y)=x^2+y^2$ and above the region in the $xy$-plane
bounded by $y=x^2$ and $x=y^2$.
Plot@8x ^ 2, Sqrt@xD<, 8x, - 1, 2<D
4
3
2
1
-1.0
-0.5
0.5
1.0
1.5
2.0
Integrate@Integrate@Integrate@1, 8z, 0, x ^ 2 + y ^ 2<D, 8y, x ^ 2, Sqrt@xD<D, 8x, 0, 1<D
6
35
Note: x^2 is smaller than Sqrt[x] on the interval [0,1]
3