finding area using double integrals The Next CEO of Stack OverflowBreaking up double integrals.Finding the double integral over a specified region.Region of integration for a double integralIntegrating over the region contained by the unit disk.Find the area of this quadrilateral using double integralsEvaluate the double integral boundedChanging the order of integration on a rectangular and polar regionswitch the double integral and evaluate it.Using double integrals to find area between two graphsReversing the order of integration to solve the double integral
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finding area using double integrals
The Next CEO of Stack OverflowBreaking up double integrals.Finding the double integral over a specified region.Region of integration for a double integralIntegrating over the region contained by the unit disk.Find the area of this quadrilateral using double integralsEvaluate the double integral boundedChanging the order of integration on a rectangular and polar regionswitch the double integral and evaluate it.Using double integrals to find area between two graphsReversing the order of integration to solve the double integral
$begingroup$
Use double integral to find the area of the region bounded by the parabola $x=y-y^2$ and the line $y=-x$
I am able find the area using $dxdy$ $A=int _0^2int _-y^y-y^2dxdy=frac43$
When reversing the order of integration, I have $A=int _-2^0int _-x^frac12+frac12sqrt1+4xdydx +int_0^frac14int _frac12-frac12sqrt1+4x^frac12+frac12sqrt1+4xdydx$
When integrating, one of the terms involves $(1+4x)^frac32$ and I can't evaluate this at $x=-2$
calculus
asked Mar 20 at 8:19
mamotebangmamotebang
152
$endgroup$
add a comment |
$begingroup$
Use double integral to find the area of the region bounded by the parabola $x=y-y^2$ and the line $y=-x$
I am able find the area using $dxdy$ $A=int _0^2int _-y^y-y^2dxdy=frac43$
When reversing the order of integration, I have $A=int _-2^0int _-x^frac12+frac12sqrt1+4xdydx +int_0^frac14int _frac12-frac12sqrt1+4x^frac12+frac12sqrt1+4xdydx$
When integrating, one of the terms involves $(1+4x)^frac32$ and I can't evaluate this at $x=-2$
calculus
asked Mar 20 at 8:19
mamotebangmamotebang
152
$endgroup$
add a comment |
$begingroup$
Use double integral to find the area of the region bounded by the parabola $x=y-y^2$ and the line $y=-x$
I am able find the area using $dxdy$ $A=int _0^2int _-y^y-y^2dxdy=frac43$
When reversing the order of integration, I have $A=int _-2^0int _-x^frac12+frac12sqrt1+4xdydx +int_0^frac14int _frac12-frac12sqrt1+4x^frac12+frac12sqrt1+4xdydx$
When integrating, one of the terms involves $(1+4x)^frac32$ and I can't evaluate this at $x=-2$
calculus
asked Mar 20 at 8:19
mamotebangmamotebang
152
$endgroup$
Use double integral to find the area of the region bounded by the parabola $x=y-y^2$ and the line $y=-x$
I am able find the area using $dxdy$ $A=int _0^2int _-y^y-y^2dxdy=frac43$
When reversing the order of integration, I have $A=int _-2^0int _-x^frac12+frac12sqrt1+4xdydx +int_0^frac14int _frac12-frac12sqrt1+4x^frac12+frac12sqrt1+4xdydx$
When integrating, one of the terms involves $(1+4x)^frac32$ and I can't evaluate this at $x=-2$
calculus
calculus
asked Mar 20 at 8:19
mamotebangmamotebang
152
asked Mar 20 at 8:19
mamotebangmamotebang
152
asked Mar 20 at 8:19
mamotebangmamotebang
152
asked Mar 20 at 8:19
mamotebangmamotebang
152
asked Mar 20 at 8:19
mamotebangmamotebang
152
152
add a comment |
add a comment |
1 Answer
1
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$begingroup$
A sign error tripped you up. The solutions of $y^2-y+x=0$ are $y=frac-1pmsqrt1-4x2$, so $sqrt1+4x$ should never have appeared. So the calculation should be $$int_-2^0left(frac12+x+frac12sqrt1-4xright)dx+int_0^1/4sqrt1-4xdx\=left[fracx2+fracx^22-frac112(1-4x)^3/2right]_-2^0+left[-frac16(1-4x)^3/2right]_0^1/4\=1-2-frac112left(1-27right)+frac16=frac43.$$
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A sign error tripped you up. The solutions of $y^2-y+x=0$ are $y=frac-1pmsqrt1-4x2$, so $sqrt1+4x$ should never have appeared. So the calculation should be $$int_-2^0left(frac12+x+frac12sqrt1-4xright)dx+int_0^1/4sqrt1-4xdx\=left[fracx2+fracx^22-frac112(1-4x)^3/2right]_-2^0+left[-frac16(1-4x)^3/2right]_0^1/4\=1-2-frac112left(1-27right)+frac16=frac43.$$
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
$endgroup$
add a comment |
$begingroup$
A sign error tripped you up. The solutions of $y^2-y+x=0$ are $y=frac-1pmsqrt1-4x2$, so $sqrt1+4x$ should never have appeared. So the calculation should be $$int_-2^0left(frac12+x+frac12sqrt1-4xright)dx+int_0^1/4sqrt1-4xdx\=left[fracx2+fracx^22-frac112(1-4x)^3/2right]_-2^0+left[-frac16(1-4x)^3/2right]_0^1/4\=1-2-frac112left(1-27right)+frac16=frac43.$$
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
$endgroup$
add a comment |
$begingroup$
A sign error tripped you up. The solutions of $y^2-y+x=0$ are $y=frac-1pmsqrt1-4x2$, so $sqrt1+4x$ should never have appeared. So the calculation should be $$int_-2^0left(frac12+x+frac12sqrt1-4xright)dx+int_0^1/4sqrt1-4xdx\=left[fracx2+fracx^22-frac112(1-4x)^3/2right]_-2^0+left[-frac16(1-4x)^3/2right]_0^1/4\=1-2-frac112left(1-27right)+frac16=frac43.$$
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
$endgroup$
A sign error tripped you up. The solutions of $y^2-y+x=0$ are $y=frac-1pmsqrt1-4x2$, so $sqrt1+4x$ should never have appeared. So the calculation should be $$int_-2^0left(frac12+x+frac12sqrt1-4xright)dx+int_0^1/4sqrt1-4xdx\=left[fracx2+fracx^22-frac112(1-4x)^3/2right]_-2^0+left[-frac16(1-4x)^3/2right]_0^1/4\=1-2-frac112left(1-27right)+frac16=frac43.$$
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
edited Mar 20 at 8:43
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
answered Mar 20 at 8:29
J.G.J.G.
32.5k23250
32.5k23250
add a comment |
add a comment |
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