Recent content by Ral
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Use polar coordinates to evaluate.
Geez, making such stupid mistakes. That part that's hard to read is e^x, sorry about that. I'll retype it here. \int\int_{R}ye^x So when I go into polar coordinates, I replace x and y with rcos\vartheta and rsin\vartheta I get \int^{\pi/2}_{0}\int^{5}_{0}rsin\vartheta*e^{rcos\vartheta}...- Ral
- Post #5
- Forum: Calculus and Beyond Homework Help
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Use polar coordinates to evaluate.
Homework Statement http://img162.imageshack.us/img162/9831/97118623.jpg Homework Equations The Attempt at a Solution I first drew R, and from the circle equation, I know the radius of the circle is 12.5. Since the region is in the first quadrant, that'll mean that my limits of...- Ral
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- Coordinates Polar Polar coordinates
- Replies: 6
- Forum: Calculus and Beyond Homework Help
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Changing a double integral to polar coordinates
Yeah, I was going to do that simplification. Thanks for helping me clarify this.- Ral
- Post #9
- Forum: Calculus and Beyond Homework Help
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Changing a double integral to polar coordinates
Yes, I mean the limits of integration for r and theta. So if that R is correct, then that means I had the correct answer, right?- Ral
- Post #7
- Forum: Calculus and Beyond Homework Help
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Changing a double integral to polar coordinates
http://img96.imageshack.us/img96/5444/66431931.jpg Here's what I had. The 1st quadrant would be shaded. The one that's confusing me the most is the integral for dr, while I think the integral for d\vartheta is correct.- Ral
- Post #5
- Forum: Calculus and Beyond Homework Help
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Changing a double integral to polar coordinates
That limit should have been \sqrt{2x-x^2. I messed up typing it up, I edited it in the right limit.- Ral
- Post #3
- Forum: Calculus and Beyond Homework Help
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Changing a double integral to polar coordinates
Homework Statement Rewrite by converting to polar coordinates, carefully drawing R. \int^{2}_{0}\int^{\sqrt{2x-x^2}}_{0}\sqrt{x^2+y^2}dydxHomework Equations The Attempt at a Solution I believe I have the inside part of it right. What I did was replace the x^2 and y^2 in \sqrt{x^2+y^2} with...- Ral
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- Coordinates Double integral Integral Polar Polar coordinates
- Replies: 8
- Forum: Calculus and Beyond Homework Help
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Finding Cartesian equation of parametric surface.
Homework Statement Find the Cartesian equation of the parametric surface: [2cos(t)cos(s), 3sin(s), sin(t)cos(s)] Find eqn. of the tangent plane when S = 0, t = pi/2Homework Equations The Attempt at a Solution I'm not quite sure what to do. All I've done is squared each term, which gave me...- Ral
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- Cartesian Parametric Surface
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Curve and tangent of a surface intersected by a plane
z = \sqrt{x^2 + 1} Then would that be the curve?- Ral
- Post #5
- Forum: Calculus and Beyond Homework Help
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Curve and tangent of a surface intersected by a plane
Yeah, z = \sqrt{x^2 + y^2} is the surface. So the curve would then be \sqrt{x^2 + 1}?- Ral
- Post #3
- Forum: Calculus and Beyond Homework Help
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Curve and tangent of a surface intersected by a plane
Homework Statement a.)\sqrt{x^2+y^2} Find the equation of the tangent plane at the point given by: x = 1, y = 1 Draw the 3d-graph of the surface and the tangent plane. \stackrel{\rightarrow}{n} = the normal vector to the tangent plane. b.) If the surface is intersected with the plane y =...- Ral
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- Curve Plane Surface Tangent
- Replies: 5
- Forum: Calculus and Beyond Homework Help