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Solve this 2 question Q1. Given a double integral $\int_{0}^{\pi} \int_{x}^{\pi...
Sep 1, 2024
Solve this 2 question
Solution by Steps
For Question 1:
step 1
To sketch the integration region RR, we need to identify the limits of integration: 0≤x≤π0 \leq x \leq \pi and x≤y≤πx \leq y \leq \pi. This describes a triangular region in the xyxy-plane
step 2
To reverse the order of integration, we express the limits for yy first: 0≤y≤π0 \leq y \leq \pi and 0≤x≤y0 \leq x \leq y. Thus, the double integral becomes I=∫0π∫0y(y−sin⁡xcos⁡x) dx dyI = \int_{0}^{\pi} \int_{0}^{y} (y - \sin x \cos x) \, dx \, dy
step 3
Evaluating the inner integral: I=∫0π[yx−12sin⁡2x]0y dy=∫0π(y2−12sin⁡2y) dyI = \int_{0}^{\pi} \left[ yx - \frac{1}{2} \sin^2 x \right]_{0}^{y} \, dy = \int_{0}^{\pi} \left( y^2 - \frac{1}{2} \sin^2 y \right) \, dy
step 4
Now, we compute the integral: I=∫0πy2 dy−12∫0πsin⁡2y dy=π33−12⋅π2=π33−π4I = \int_{0}^{\pi} y^2 \, dy - \frac{1}{2} \int_{0}^{\pi} \sin^2 y \, dy = \frac{\pi^3}{3} - \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi^3}{3} - \frac{\pi}{4}
step 5
Finally, simplifying gives us the result: I≈9.55I \approx 9.55
Answer
9.55
Key Concept
Reversing the order of integration in double integrals
Explanation
This process allows us to evaluate the integral more easily by changing the limits of integration based on the region of integration.
--- For Question 2:
step 1
To find the volume of the solid enclosed by the cylinders y=x2y = x^2 and x=y2x = y^2, we first determine the intersection points of these curves, which are (0,0)(0,0) and (1,1)(1,1)
step 2
The volume can be expressed as a double integral: V=∫01∫x2x(2−x−y) dy dxV = \int_{0}^{1} \int_{x^2}^{\sqrt{x}} (2 - x - y) \, dy \, dx
step 3
Evaluating the inner integral: V=∫01[(2−x)y−y22]y=x2y=x dxV = \int_{0}^{1} \left[ (2 - x)y - \frac{y^2}{2} \right]_{y=x^2}^{y=\sqrt{x}} \, dx
step 4
Substituting the limits gives us: V=∫01[(2−x)x−x2] dxV = \int_{0}^{1} \left[ (2 - x)\sqrt{x} - \frac{x}{2} \right] \, dx
step 5
Finally, evaluating this integral results in V=1130V = \frac{11}{30}
Answer
1130\frac{11}{30}
Key Concept
Using double integrals to find volumes of solids
Explanation
This method allows us to calculate the volume of a solid by integrating over the area defined by the curves and planes.
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Solution by Steps
For Q3: Evaluate the triple integral
step 1
We set up the triple integral as follows: I=∫01∫0x∫0x+y(x+y+z) dz dy dxI = \int_{0}^{1} \int_{0}^{x} \int_{0}^{x+y} (x+y+z) \, dz \, dy \, dx
step 2
First, we evaluate the innermost integral: ∫0x+y(x+y+z) dz=(x+y)z+z22∣0x+y=(x+y)(x+y)+(x+y)22=3(x+y)22\int_{0}^{x+y} (x+y+z) \, dz = (x+y)z + \frac{z^2}{2} \bigg|_{0}^{x+y} = (x+y)(x+y) + \frac{(x+y)^2}{2} = \frac{3(x+y)^2}{2}
step 3
Now, we substitute this result into the next integral: I=∫01∫0x3(x+y)22 dy dxI = \int_{0}^{1} \int_{0}^{x} \frac{3(x+y)^2}{2} \, dy \, dx
step 4
Evaluating the integral with respect to yy: I=∫0132[(x+y)33]0x dx=∫0132((2x)33−x33) dx=∫0132(8x3−x33) dx=∫017x32 dxI = \int_{0}^{1} \frac{3}{2} \left[ \frac{(x+y)^3}{3} \right]_{0}^{x} \, dx = \int_{0}^{1} \frac{3}{2} \left( \frac{(2x)^3}{3} - \frac{x^3}{3} \right) \, dx = \int_{0}^{1} \frac{3}{2} \left( \frac{8x^3 - x^3}{3} \right) \, dx = \int_{0}^{1} \frac{7x^3}{2} \, dx
step 5
Finally, we evaluate the last integral: I=72⋅x44∣01=78I = \frac{7}{2} \cdot \frac{x^4}{4} \bigg|_{0}^{1} = \frac{7}{8}
Answer
7/87/8
Key Concept
Triple integrals allow us to compute volumes and other quantities in three-dimensional space.
Explanation
The evaluation of the triple integral involves integrating step-by-step, starting from the innermost integral and working outward, which simplifies the computation.
--- For Q4: Use a triple integral to find the volume of the solid enclosed by the paraboloids
step 1
The volume VV is given by the triple integral: V=∫−22∫04−y2∫018−x2−y2dz dx dyV = \int_{-2}^{2} \int_{0}^{\sqrt{4-y^2}} \int_{0}^{18-x^2-y^2} dz \, dx \, dy
step 2
We first evaluate the innermost integral: ∫018−x2−y2dz=18−x2−y2\int_{0}^{18-x^2-y^2} dz = 18 - x^2 - y^2
step 3
Substituting this into the next integral gives: V=∫−22∫04−y2(18−x2−y2) dx dyV = \int_{-2}^{2} \int_{0}^{\sqrt{4-y^2}} (18 - x^2 - y^2) \, dx \, dy
step 4
Now we evaluate the integral with respect to xx: V=∫−22[18x−x33−y2x]04−y2 dyV = \int_{-2}^{2} \left[ 18x - \frac{x^3}{3} - y^2x \right]_{0}^{\sqrt{4-y^2}} \, dy
step 5
Evaluating this results in V=∫−22(184−y2−(4−y2)3/23−y24−y2) dyV = \int_{-2}^{2} \left( 18\sqrt{4-y^2} - \frac{(4-y^2)^{3/2}}{3} - y^2\sqrt{4-y^2} \right) \, dy which simplifies to 81π81\pi after integration
Answer
81π81\pi
Key Concept
Triple integrals can be used to find volumes of solids in three-dimensional space defined by surfaces.
Explanation
The volume is calculated by integrating the height of the solid over the base area defined by the intersection of the paraboloids.
--- For Q5: Find the volume of the solid bounded by the plane $z=y+2$ and the cylinder $x^{2}+y^{2}=4$ above the $xy$-plane.
step 1
The volume VV can be expressed as: V=∫−22∫−4−x24−x2(y+2) dy dxV = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (y+2) \, dy \, dx
step 2
We first evaluate the inner integral: ∫−4−x24−x2(y+2) dy=[y22+2y]−4−x24−x2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (y+2) \, dy = \left[ \frac{y^2}{2} + 2y \right]_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}
step 3
Evaluating this gives: =((4−x2)2+24−x2)−((4−x2)2−24−x2)=44−x2= \left( \frac{(4-x^2)}{2} + 2\sqrt{4-x^2} \right) - \left( \frac{(4-x^2)}{2} - 2\sqrt{4-x^2} \right) = 4\sqrt{4-x^2}
step 4
Now substituting this into the outer integral gives: V=∫−2244−x2 dxV = \int_{-2}^{2} 4\sqrt{4-x^2} \, dx
step 5
Evaluating this integral results in V=8πV = 8\pi after applying the appropriate trigonometric substitution
Answer
8π8\pi
Key Concept
The volume of solids can be calculated using double integrals over regions defined by curves and surfaces.
Explanation
The volume is determined by integrating the height of the solid above the base area defined by the cylinder.
--- For Q6: Find the parametric equations for a smooth curve, $C: x=5+z, y=3z^{2}$ that joins the point from $P(6,3,1)$ to $Q(5,0,0)$.
step 1
The parametric equations are given as: x=5+t,y=3t2,z=tx = 5 + t, y = 3t^2, z = t where tt varies from 11 to 00
step 2
At t=1t=1, we have the point P(6,3,1)P(6,3,1): x=5+1=6,y=3(1)2=3,z=1x = 5 + 1 = 6, y = 3(1)^2 = 3, z = 1
step 3
At t=0t=0, we have the point Q(5,0,0)Q(5,0,0): x=5+0=5,y=3(0)2=0,z=0x = 5 + 0 = 5, y = 3(0)^2 = 0, z = 0
step 4
Thus, the parametric equations correctly describe the curve from PP to QQ
Answer
x=5+t,y=3t2,z=tx=5+t, y=3t^{2}, z=t from t=1t=1 to t=0t=0
Key Concept
Parametric equations represent curves in space by expressing coordinates as functions of a parameter.
Explanation
The equations describe a smooth transition from point PP to point QQ in three-dimensional space.
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Solution by Steps
For Q11:
step 1
To apply Green's theorem, we need to compute the double integral of the curl of the vector field over the region enclosed by the curve C C . The vector field is given by P=5xy+x2+y2 P = 5xy + x^2 + y^2 and Q=x2−y Q = x^2 - y
step 2
The curl is calculated as follows: ∂Q∂x−∂P∂y=∂(x2−y)∂x−∂(5xy+x2+y2)∂y=2x−(5x+2y)=−3x−2y \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial (x^2 - y)}{\partial x} - \frac{\partial (5xy + x^2 + y^2)}{\partial y} = 2x - (5x + 2y) = -3x - 2y
step 3
The region R R is bounded by y=x y = x , x2+y2=4 x^2 + y^2 = 4 , and the y-axis. We need to set up the double integral: ∬R(−3x−2y) dA \iint_R (-3x - 2y) \, dA
step 4
The limits of integration can be determined from the curves. The integral becomes ∫02∫04−x2(−3x−2y) dy dx \int_0^2 \int_0^{\sqrt{4 - x^2}} (-3x - 2y) \, dy \, dx
step 5
Evaluating this integral gives us the result −6.11 -6.11
Answer
-6.11
--- For Q12:
step 1
We apply Green's theorem to the line integral ∫Cxy dx+x2 dy \int_C xy \, dx + x^2 \, dy . The vector field is P=xy P = xy and Q=x2 Q = x^2
step 2
The curl is calculated as ∂Q∂x−∂P∂y=2x−x=x \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - x = x
step 3
The region R R is bounded by the parabola y=9−x2 y = 9 - x^2 and the x-axis. We set up the double integral: ∬Rx dA \iint_R x \, dA
step 4
The limits of integration are 0≤x≤3 0 \leq x \leq 3 and 0≤y≤9−x2 0 \leq y \leq 9 - x^2 . The integral becomes ∫03∫09−x2x dy dx \int_0^3 \int_0^{9 - x^2} x \, dy \, dx
step 5
Evaluating this integral results in 0 0
Answer
0
--- For Q13:
step 1
We apply Green's theorem to the line integral ∮C(x−3y2) dx+2(3+xy) dy \oint_C (x - 3y^2) \, dx + 2(3 + xy) \, dy . The vector field is P=x−3y2 P = x - 3y^2 and Q=2(3+xy) Q = 2(3 + xy)
step 2
The curl is calculated as ∂Q∂x−∂P∂y=2y−(−6y)=2y+6y=8y \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - (-6y) = 2y + 6y = 8y
step 3
The region R R is bounded by the triangle with vertices A(1,0) A(1,0) , B(1,6) B(1,6) , and C(7,6) C(7,6) . We set up the double integral: ∬R8y dA \iint_R 8y \, dA
step 4
The limits of integration can be determined from the vertices. The integral becomes ∫17∫068y dy dx \int_1^7 \int_0^6 8y \, dy \, dx
step 5
Evaluating this integral gives us the result 576 576
Answer
576
---
Key Concept
Green's theorem relates a line integral around a simple closed curve to a double integral over the region it encloses.
Explanation
The answers to the problems demonstrate the application of Green's theorem to evaluate line integrals by converting them into double integrals over the corresponding regions.
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